# Resolução – Física I – Mecânica – Sears; Zemansky

Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch.

?1000cm3 ? ? 1in ?3 1.2: 0.473 L ?ù ? ? ?ù ? ? = 28.9 in 3 . ? 1 L ? ? 2.54 cm ?

1.3: The time required for light to travel any distance in a vacuum is the distance divided by the speed of light;

3.00 ?ù108 m s g ? 1kg ? ?100cm?3 kg 1.4: 11.3 ?ù? ??ù? ? =1.13 ?ù 104 . cm3 ?1000 g ? ? 1 m ? m3

3 ? 1000 L ? ? 1 gal ? ? 128 oz. ? ? 1 bottle ? 1.6: 1 m3 ?ù ? 3 ? ?ù ? ? ?ù ? ??ù ? ?. ? 1 m ? ? 3.788 L ? ? 1 gal ? ? 16 oz. ? = 2111.9 bottles ? 2112 bottles The daily consumption must then be bottles ? 1 yr ? bottles yr ? 365.24 da ? da

furlongs ? 1 mile ? ? 1 fortnight ? ? 1 day ? mi 1.8: 180,000 ?ù ? ? ?ù ? ? ?ù ? ? = 67 . fortnight ? 8 furlongs ? ? 14 day ? ? 24 h ? h

? mi ? ? 1h ? ? 5280 ft ? ft 1.10: a) ? 60 ? ? ? ? ? = 88 ? hr ? ? 3600 s ? ? 1mi ? s ? ft ? ? 30.48 cm ? ? 1 m ? m b) ? 32 ? ? ? ? ? = 9.8 ? s 2 ? ? 1ft ? ? 100 cm ? s 2

? g??100cm?3?1kg? 3kg c) ?1.0 ? ? ? ? ? = 10 ? cm3?? 1m ? ?1000g? m3

1.11: The density is mass per unit volume, so the volume is mass divided by density. V=(60?ù103g)(19.5gcm3)=3077cm3 4 Use the formula for the volume of a sphere, V = ?r 3 , 3 to calculate r : r = (3V 4? ) = 9.0 cm 1/3

1.12: (3.16?ù107s-??ù107s)(3.16?ù107s)?ù100=0.58%

10 m – 890 ?ù103 m b) Since the distance was given as 890 km, the total distance should be 890,000 To report the total distance as 890,010 meters, the distance should be given as 890.01 km.

0.13%. b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3?ù10-3%. c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about 2.8?ù10-2%.

The area is 9.69 ?? 0.07 cm2, where the extreme values in the piece’s length and 1.16: width are used to find the uncertainty in the area. The fractional uncertainty in the area is 0.07 cm2 = 0.72%, and the fractional uncertainties in the length and width are 9.69 cm 5.10 cm 1.9 cm

1.17: a) The average volume is ? (8.50 cm) (0.050 cm) = 2.8 cm3 2

4 (two significant figures) and the uncertainty in the volume, found from the extreme values of the diameter and thickness, is about 0.3 cm3 , and so the volume of a cookie is 2.8 ?? 0.3 cm3. (This method does not use the usual form for progation of errors, which is not addressed in the text. The fractional uncertainty in the thickness is so much greater than the fractional uncertainty in the diameter that the fractional uncertainty in the volume is 10% , reflected in the above answer.)

.05 1.18: (Number of cars ?ù miles/car.day)/mi/gal = gallons/day (2 ?ù 108 cars ?ù 10000 mi/yr/car ?ù 1 yr/365 days)/(20 mi/gal) = 2.75 ?ù 108 gal/day

1.19: Ten thousand; if it were to contain ten million, each sheet would be on the order of a millionth of an inch thick.

1.20: If it takes about four kernels to fill 1 cm3, a 2-L bottle will hold about 8000 kernels.

each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercise and problems), so an estimate for the number of words is about 106 .

1.22: Assuming about 10 breaths per minutes, 24 ?ù 60 minutes per day, 365 days per year, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime isabout2?ù105m3.Thisisthevolumeofaroom100m?ù100m?ù20m,whichiskindof tight for a major-league baseball game, but it’s the same order of magnitude as the volume of the Astrodome.

1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 105 times per day. With 365 days in a year and the above lifespan of 80 years, the number of beatsinalifetimeisabout3?ù109.With1L(50cm3)perbeat,andabout1gallonper 20 4 liter, this comes to about 4 ?ù107 gallons.

1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be in the shape of a pyramid, say with a height of 2 m and a base 3 m ?ù 3 m . The volume of such a pile is 6 m3 , and the calculations of Example 1-4 indicate that the value of this volume is $6 ?ù108.

1.26: Thesurfaceareaoftheearthisabout4?R2=5?ù1014m2,whereRistheradiusof theearth,about6?ù106m,sothesurfaceareaofalltheoceansisabout4?ù1014m2.An averagedepthofabout10kmgivesavolumeof4?ù1018m3=4?ù1024cm3.Characterizing the size of a ÔÇ£dropÔÇØ is a personal matter, but 25 drops cm3 is reasonable, giving a total of 1026 drops of water in the oceans.

1.27: This will of course depend on the size of the school and who is considered a “student”. A school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) will total 104 pizzas, as will a school of 250 students averaging 40 pizzas a year each.

bills can be stacked with about 2-3 per millimeter, so the number of bills in a stack to the moon would be about 1012. The value of these bills would be $1 trillion (1 terabuck).

(9,372,571km2?ù106m2km2)(15.6cm?ù6.7cm?ù1m2104cm2)=9?ù1014bills 1.30:

a) 11.1 m @ 77.6o b) 28.5 m @ 202o c) 11.1 m @ 258o d) 28.5 m @ 22o

1.35: r A = (12.0 m)sin 37.0o = 7.2 m, A = (12.0 m) cos 37.0o = 9.6 m. xy r = (15.0 m)cos 40.0o = = -(15.0 m) o = -9.6 m. B; B 11.5 m, B sin 40.0 xy

r (6.0 m) (6.0 m) xy A -1.00 m (a) tan ? = y = = -0.500 1.36: A 2.00 m X

?=tan-1(-0.500)=360o-26.6o=333o A 1.00 m (b) tan ? = y = = 0.500 A 2.00 m x

? = tan -1 (0.500) = 26.6o A 1.00 m (c) tan ? = y = = -0.500 A – 2.00 m x

? = tan (- 0.500) = 180o – 26.6o = 153o -1 A – 1.00 m (d) tan ? = y = = 0.500 A – 2.00 m x

second force has components F = F cos 32.4o = 433 N and F = F sin 32.4o = 275 N. 2x 2 2y 2 1x 1y F = F + F = 1158 N and F = F + F = 275 N x 1x 2 x y 1 y 2 y

r 1.39: Using components as a check for any graphical method, the components of B are r xy x a) The x – and y – components of the sum are 2.4 m and 10.8 m, for a magnitude ( )2 ( )2 ? 10.8 ? ? 2.4 ? c) The x- and y-components of the vector difference are ÔÇô 26.4 m and – 10.8 m, for a magnitude of 28.5 m and a direction arctan ( -10.8 ) = 202o. Note that -26.4 180o must be added to arctan( -10.8 ) = arctan(10.8 ) = 22o in order to give an angle in the – 26.4 26.4 third quadrant.

rr j (26.4 )2 (10 )2 ? 10.8 ? Magnitude = m + .8 m = 28.5 m at and angle of arctan? ? = 22.2o. ? 26.4 ?

1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given vectors is:

a) (-8.6 cm)2 + (5.20 cm)2 = 10.0 cm, arctan ( 5.20 ) = 148.8o (which is -8.60 180o ÔÇô 31.2o).

The total northward displacement is 3.25 km – 1.50 km = 1.75 km, , and the total westward displacement is 4.75 km . The magnitude of the net displacement is (1.75 km)2 + (4.75 km)2 = 5.06 km. The south and west displacements are the same, so The direction of the net displacement is 69.80o West of North.

1.42: a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + (ÔÇô3.75 cm) = ÔÇô1.50 cm.

rr 1.43: a) The magnitude of A + B is ? ((2.80 ) + (1.90 ) )2 ? cm cos 60.0o cm cos 60.0o ? ? = 2.48 cm ? ((2.80 ) (1.90 ) )2 ? ? + cm sin 60.0o – cm sin 60.0o ? and the angle is ?(2.80cm)sin60.0o-(1.90cm)sin60.0o? arctan ? (2.80 ) (1.90 )cos 60.0o ? = 18 o

? cm cos 60.0o + cm ? rr b) The magnitude of A – B is

? ((2.80 cm) cos 60.0 – (1.90 cm) cos 60.0 )2 ? oo ? ? = 4.10 cm + ((2.80 cm) sin 60.0o + (1.90 cm) sin 60.0o )2 ?? ?? and the angle is ? (2.80 cm) sin 60.0o + (1.90 cm)sin 60.0o ? arctan ? (2.80 cm) cos 60.0o – (1.90 cm)cos 60.0o ? = 84 o

?? r r (r r) ? 1.44: A = (ÔÇô12.0 m) i^ . More precisely,

a) Ar = (3.60 m)cos 70.0o ^ + (3.60 m)sin 70.0o ^ = (1.23 m)i^ + (3.38 m) ^j 1.46: i j Br=-(2.40m)cos30.0oi^-(2.40m)sin30.0o^j=(-2.08m)i^+(-1.20m)^j b) = (3.00) r – (4.00) r r CAB =(3.00)(1.23m)i^+(3.00)(3.38m)^j-(4.00)(-2.08m)i^-(4.00)(-1.20m)^j =(12.01m)i^+(14.94)^j

(Note that in adding components, the fourth figure becomes significant.) c) From Equations (1.8) and (1.9), ( )2 ( )2 ? 14.94 m ? C = 12.01 m + 14.94 m = 19.17 m, arctan ? ? = 51.2o ? 12.01 m ?

= (4.00)2 + (3.00)2 = = (5.00)2 + (2.00)2 = 1.47: a) A 5.00, B 5.39

= 12 + 12 + 12 = 3 ? 1 so it is not a unit vector b) r A = A2 + A2 + A2 xyz

r If any component is greater than + 1 or less than ÔÇô1, A ? 1 , so it cannot be a unit r vector. A can have negative components since the minus sign goes away when the component is squared.

c) r A =1 2 (3.0)2 + 2 (4.0)2 = aa1 a 2 25 = 1 1 a = ?? = ??0.20 5.0

rr xy xy r r ( )i^ (A ) ^j A+B=A+B + +B xxyy

r r ( )i^ (B ) ^j B+A=B+A + +A xxyy rrrr Scalar addition is commutative, so A + B = B + A

rr A?B=AB+AB xx yy rr B?A=BA+BA xx yy rrrr Scalar multiplication is commutative, so A ? B = B ? A

b) r?ùr=(AB-AB)^+(AB-AB)^+(AB-AB)^ AB i j k yz zy zx xz xy yx

r ?ù r ( )^ ( ) ^ ( ) ^ BA=BA-BAi+BA-BAj+BA-BAk yz zy zx xz xy yx Comparison of each component in each vector product shows that one is the negative of the other.

ABcos?=(12m?ù15m)cos93o=-9.4m2 BCcos?=(15m?ù6m)cos80o=15.6m2 ACcos?=(12m?ù6m)cos187o=-71.5m2

Method 2: (Sum of products of components) A?B=(7.22)(11.49)+(9.58)(-9.64)=-9.4m2 B ? C = (11.49)(-3.0) + (-9.64)(-5.20) = 15.6 m2 A ? C = (7.22)(-3.0) + (9.58)(-5.20) = -71.5 m2

1.51: a) From Eq.(1.21), r r (4.00)(5.00) (3.00)(- ) A ? B = + 2.00 = 14.00.

1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle ? as rr ?A?B? ?AB+AB? ? AB ? ? AB ? In the intermediate calculations given here, the significant figures in the dot products and in the magnitudes of the vectors are suppressed.

rr a) A?B=-22,A=40,B=13,andso ? – 22 ? ? 40 13 ?

r r ? 60 ? ? 34 136 ? rr c) A ? B = 0, ? = 90.

1.54: a) From Eq. (1.22), the magnitude of the cross product is

(12.0 m)(18.0 m)sin (180o – 37o ) = 130 m2 The right-hand rule gives the direction as being into the page, or the ÔÇô z-direction. Using Eq. (1.27), the only non-vanishing component of the cross product is C=AB=(-12m)((18.0m)sin37o)=-130m2 z xy

b) The same method used in part (a) can be used, but the relation given in Eq. (1.23) gives the result directly: same magnitude (130 m2), but the opposite direction (+z-direction).

1.55: In Eq. (1.27), the only non-vanishing component of the cross product is

C = A B – A B = (4.00)(- 2.00) – (3.00)(5.00) = -23.00, z xy yx

rr 1.56: a) From the right-hand rule, the direction of A?ù B is into the page (the ÔÇô z-direction). The magnitude of the vector product is, from Eq. (1.22), Or, using Eq. (1.27) and noting that the only non-vanishing component is C=AB-AB z xy yx

= (2.80 cm)cos60.0o (-1.90 cm)sin 60o – (2.80 cm)sin 60.0o (1.90 cm)cos 60.0o = -4.61 cm2 gives the same result.

b) Rather than repeat the calculations, Eq. (1-23) may be used to see that rr ?ù has magnitude 4.61 cm2 and is in the +z-direction (out of the page). BA

1.57: a) The area of one acre is 1 mi ?ù 1 mi = 1 mi2 , so there are 640 acres to a square 8 80 640 mile.

? 1mi2 ? ?5280ft?2 b) (1 acre)?ù ? ? ?ù ? ? = 43,560 ft 2 ? 640 acre ? ? 1 mi ? (all of the above conversions are exact).

(43,560 ) ? 7.477 gal ? c) (1acre-foot)= ft3?ù? ?=3.26?ù105gal, ? 1 ft 3 ? which is rounded to three significant figures.

1.58:a) ($4,950,000102acres)?ù(1acre43560ft2)?ù(10.77ft2m2)=$12m2. c) $.008in2?ù(1in?ù78in)=$.007forpostagestampsizedparcel.

1.420 ?ù109 Hz ? cycles ? ? 3600 s ? cycles b) ?1.420 ?ù109 ? ?ù ? ? = 5.11?ù1012 ? s ? ? 1h ? h

c) Using the conversion from years to seconds given in Appendix F, ( ) ? 3.156 ?ù107 s ? ( ) ? 1y ?

1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to find the mass of one H O molecule: 18.015 u ?ù 1.661 ?ù 10ÔÇô27 kg/u = 2.992 ?ù 10ÔÇô26 2 kg/molecule. (70 kg/2.992 ?ù 10ÔÇô26 kg/molecule) = 2.34 ?ù 1027 molecules. (Assuming carbon to be the most common atom gives 3 ?ù 1027 molecules.

1.61: a) Estimate the volume as that of a sphere of diameter 10 cm: = 4 ? 3 = ?ù -4 3 V r 5.2 10 m 3 Mass is density times volume, and the density of water is 1000 kg m3 , so m=(0.98)(1000kg m3)(5.2?ù10-4m3)=0.5kg

b) Approximate as a sphere of radius r = 0.25?Ám (probably an over estimate) = 4 ? 3 = ?ù -20 3 V r 6.5 10 m 3 m=(0.98)(1000kg m3)(6.5?ù10-20m3)=6?ù10-17kg=6?ù10-14g

MM 1.62: a) ? = , so V = V? 0.200 kg x3= =2.54?ù10-5m3 7.86?ù103 kg/m3 x=2.94?ù10-2m=2.94cm

4?R3=2.54?ù105m3 – b) 3 R=1.82?ù10-2m=1.82cm

1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume 2 more hours for restorative work. Assuming most dentists work less than 2000 hours per year, this gives 2000 hours 4 hours per patient = 500 patients per dentist. Assuming only half of the people who should go to a dentist do, there should be about 1 dentist per 1000 inhabitants. Note: A dental assistant in an office with more than one treatment room could increase the number of patients seen in a single dental office.

? 6.0 ?ù 1023 atoms ? ? 14 ?ù 10-3 kg ? mole b) The number of neutrons is the mass of the neutron star divided by the mass of a neutron: (2) (2.0?ù 1030 kg) (1.7 ?ù 10-27 kg neutron) c) The average mass of a particle is essentially 2 the mass of either the proton or 3

the neutron, 1.7 ?ù10-27 kg. The total number of particles is the total mass divided by this average, and the total mass is the volume times the average density. Denoting the density by ? (the notation introduced in Chapter 14).

4 ? R3? M (2?)(1.5?ù1011m)3(1018kgm3) = 3 = = 1.2 ?ù 1079. m 2 (1.7 ?ù 10-27 kg) ave m p 3

r r r r r r (r r r) A + B + C + D = 0, so D = – A + B + C A=+Acos30.0o=+86.6N,A=+Acos30.0o=+50.00N xy

B=-Bsin30.0o=-40.00N,B=+Bcos30.0o=+69.28N xy C=+Ccos53.0o=-24.07N,C=-Csin53.0o=-31.90N xy Then D = -22.53 N, D = -87.34 N xy

xy tan ? = D / D yx ? = 75.54o ? = 180o + ? = 256o , counterclockwise from + x – axis = 87.34 / 22.53

1.66: R = A + B = (170 km) sin 68o + (230 km) cos 48o = 311.5 km xxx R = A + B = (170 km) cos 68o – (230 km) sin 48o = -107.2 km yyy

R=R2+R2=(311.5km)2+(-107.2km)2=330km xy tan? = R R y 107.2 km = = 0.344 R x311.5 km

rrr r b) Algebraically, A = C – B, and so the components of A are A = C – B = (6.40 cm) cos 22.0o – (6.40 cm) cos 63.0o = 3.03 cm xxx yyy

xxxx =(12.0m)cos(90o-37o)+(15.00m)cos(-40o)+(6.0m)cos(180o+60o) = 15.7 m, and R=A+B+C yyyy

=(12.0m)sin(90o-37o)+(15.00m)sin(-40o)+(6.0m)sin(180o+60o) The magnitude of the resultant is R = R2 + R2 = 16.6 m , and the direction from xy

thepositivex-axisisarctan(-5.3)=-18.6o.Keepingextrasignificantfiguresinthe 15.7 intermediate calculations gives an angle of – 18.49??, which when considered as a positive counterclockwise angle from the positive x-axis and rounded to the nearest degree is 342o .

Take the east direction to be the x – direction and the north direction to be the y – direction. The x- and y-components of the resultant displacement of the first three displacements are then (-180m)+(210m)sin45o+(280m)sin30o=108m, -(210m)cos45o+(280m)cos30o=+94.0m,

keeping an extra significant figure. The magnitude and direction of this net displacement are ( )2 ( )2 ? 94 m ? ? 108 m ? The fourth displacement must then be 144 m in a direction 40.9o south of west.

The third leg must have taken the sailor east a distance (5.80 km) – (3.50 km) cos 45o – (2.00 km) = 1.33 km and a distance north (3.5 km)sin 45o = (2.47 km) The magnitude of the displacement is

(1.33 km)2 + (2.47 km)2 = 2.81 km andthedirectionisarctan(2.47)=62??northofeast,whichis90??-62??=28??east 1.33 of north. A more precise answer will require retaining extra significant figures in the intermediate calculations.

1.71: a) b) The net east displacement is -(2.80km)sin45o+(7.40km)cos30o-(3.30km)cos22o=1.37km,andthenetnorth displacementis-(2.80km)cos45o+(7.40km)sin30o-(3.30km)sin22.0o=0.48km, and so the distance traveled is (1.37 km)2 + (0.48 km)2 = 1.45 km.

(147 km)sin 85o + (106 km) sin 167o + (166 km)sin 235o = 34.3 km and the northward displacement is (147km)cos85o+(106km)cos167o+(166km)cos235o=-185.7km (A negative northward displacement is a southward displacement, as indicated in Fig. (1.33). Extra figures have been kept in the intermediate calculations.)

a) (34.3 km)2 + (185.7 km)2 = 189 km b) The direction from Lincoln to Manhattan, relative to the north, is

210 -10 second line is 42o + 30o = 72o. Therefore X = 10 + 250 cos 72o = 87 Y = 20 + 250 sin 72o = 258 b) The computer screen now looks something like this:

The length of the bottom line is (210 – 87)2 + (200 – 258) = 136 and its direction is 2

b) To use the method of components, let the east direction be the x-direction and the north direction be the y-direction. Then, the explorer’s net x- displacement is, in units of his step size, (40)cos 45o – (80)cos 60o = -11.7 and the y-displacement is (40)sin 45o + (80)sin 60o – 50 = 47.6.

The magnitude and direction of the displacement are ? 47.6 ? ? – 11.7 ? (More precision in the angle is not warranted, as the given measurements are to the nearest degree.) To return to the hut, the explorer must take 49 steps in a direction 104o – 90o = 14o east of south.

r 1.75: Let +x be east and +y be north. Let A be the displacement 285 km at 40.0o north r of west and let B be the unknown displacement.

rrr r A + B = R, where R = 115 km, east rrr B=R-A B =R -A B =R -A x x x, y y y

A=-Acos40.0o=-218.3km,A=+Asin40.0o=+183.2km xy R = 115 km, R = 0 xy xy

xy tan?=B B =(183.2km)(333.3km) yx ? = 28.8o , south of east

1.76: (a) ? = ? sin? par (b) ? = ? cos? perp (c) ? = ? sin? par

r r rrr E=R-B,E=R-B xxxyyy B=-Bsin43o=-158.2N,B=+Bcos43o=+169.7N xy R = 0, R = +132.5 N xy Then E = +158.2 N, E = -37.2N xy

direction, east the x-direction, and the z-axis vertical. The first displacement is then – 30k^, the second is – 15 ^j , the third is 200i^ (0.2 km = 200 m) , and the fourth is 100 ^j . Adding the four: – 30k^ – 15 ^j + 200i^ + 100 ^j = 200i^ + 85 ^j – 30k^

(b) The total distance traveled is the sum of the distances of the individual segments: 30 + 15 + 200 + 100 = 345 m. The magnitude of the total displacement is:

= 2 + 2 + 2 = 2 + 2 + (- )2 = D D D D 200 85 30 219 m xyz

r A = 240 m, 32o south of east rr B is 32o south of west and C is 62o south of west Let + x be east and + y be north rrr A+B+C=0 A + B + C = 0, so A cos 32o – B cos 48o – C cos 62o = 0 xxx A + B + C = 0, so – A sin 32o + B sin 48o – C sin 62o = 0 yyy A is known so we have two equations in the two unknowns B and C. Solving gives B = 255 m and C = 70 m.

1.80: Take your tent’s position as the origin. The displacement vector for Joe’s tent is (21cos23)^-(21sin23)^=19.33^-8.205^.ThedisplacementvectorforKarl’stentis oi oj i j (32cos37)^+(32sin37)^=25.56^+19.26^.Thedifferencebetweenthetwo oi oj i j displacements is: (19.33 – 25.56)i^ + (- 8.205 – 19.25) ^j = -6.23i^ – 27.46 ^j .

The magnitude of this vector is the distance between the two tents:

zz AB+AB=(Acos?)(Bcos?)+(Asin?)(Bsin?) xx yy A B A B = AB(cos ? cos ? + sin ? sin ? ) ABAB = AB cos( ? – ? ) AB = AB cos ? where the expression for the cosine of the difference between two angles has been used (see Appendix B).

r b) With A = B = 0, C = C ^ and C = C . From Eq. (1.27), k zz z z

C=AB-AB xx yx =(Acos?)(Bcos?)-(Asin?)(Bcos?) ABAA = AB cos ? sin ? – sin ? cos ? ABAB = AB sin(? – ? ) BA = AB sin ?

1.82: a) The angle between the vectors is 210o – 70o = 140o , and so Eq. (1.18) gives r r ( )( ) A?B=3.60m2.40mcos140o=-6.62m2Or,Eq.(1.21)gives rr A?B=AB+AB xx yy

=(3.60m)cos70o(2.4m)cos210o+(3.6m)sin70o(2.4m)sin210o = -6.62 m2

b) From Eq. (1.22), the magnitude of the cross product is (3.60m)(2.40m)sin140o=5.55m2,

and the direction, from the right-hand rule, is out of the page (the rr +z-direction). From Eq. (1-30), with the z-components of A and B vanishing, the z-component of the cross product is A B – A B = (3.60 m)cos 70o (2.40 m)sin 210o xy yx

Trianglearea=12(base)(height)=12(B)(Asin?) Parellogram area = BA sin ? b) 90o

1.84: With the +x-axis to the right, +y-axis toward the top of the page, and +z-axis out (r r) (r r) (r r) xyz

2 Ar – Br = (A – B )i^ + (A – B )^ + (A – B )k^ j xxyyzz b) = (- 5.00)i^ + (2.00) ^j + (7.00)k^

(5.00)2 + (2.00)2 + (7.00)2 = c) 8.83, rr and this will be the magnitude of B – A as well.

1.86: The direction vectors each have magnitude 3 , and their dot product is (1) (1) + (1) (ÔÇô1) + (1) (ÔÇô1) = ÔÇô1, so from Eq. (1-18) the angle between the bonds is arccos o

a restatement of the law of cosines. We know that C2=A2+B2+2ABcos?, rr where ? is the angle between A and B .

rr a)IfC2=A2+B2,cos?=0,andtheanglebetweenAandBis90o(thevectorsare perpendicular).

derive it. The most straightforward way, using vector algebra, is to assume the linearity of the dot product (a point used, but not explicitly mentioned in the text) to show that rr the square of the magnitude of the sum A + B is

( r r )? ( r r ) r r r r r r r r A+B A+B=A?A+A?B+B?A+B?B rrrrrr =A?A+B?B+2A?B rr = A2 + B2 + 2A ? B = A2 + B2 + 2AB cos?

Using components, if the vectors make angles ?A and ?B with the x-axis, the components of the vector sum are A cos ?A + B cos ?B and A sin ?A + B sin ?B, and the square of the magnitude is ( + )2 + (A sin ? + )2 Acos? Bcos? Bsin? A B A2(cos2A? B ) (cos2? + ) = +sin2? +B2 sin2? AABB + 2 AB(cos? cos ? + sin ? sin ? ) ABAB =A2+B2+2ABcos(?-?) AB =A2+B2+2ABcos? where ? = ?A ÔÇô ?B is the angle between the vectors.

rr rr b) A geometric consideration shows that the vectors A, B and the sum A + B rr must be the sides of an equilateral triangle. The angle between A, and B is 120o, since one vector must shift to add head-to-tail. Using the result of part (a), with A = B, the condition is that A2 = A2 + A2 + 2 A2cos ? , which solves 2

rr r to c and a to d as B,C, and D . In terms of unit vectors, rr()r() CjkDijk

Using Eq. (1.18), rr ? B ? D ? ? L2 ? arccos? ?=arccos?(L)(L )?=54.7o, ? BD ? ? 3 ? rr ? C ? D ? ? 2L2 ? ? ? ? ? ? 3?

1.90: From Eq. (1.27), the cross product is ? ? 6.00 ? 11.00 ? (-13.00) i^ + (6.00) ^j + (-11.00) k^ = 13 ?- (1.00 ) i^ + ? ? ^j – k^ ?. ? ? 13.00 ? 13.00 ?

The magnitude of the vector in square brackets is 1.93, and so a unit vector in this rr direction (which is necessarily perpendicular to both A and B) is

? – (1.00) i^ + (6.00 13.00) ^j – (11.00 13) ^ ? k ? 1.93 ?

The negative of this vector, ? (1.00) i^ – (6.00 13.00) ^j + (11.00 13) ^ ? k ? ?, ? 1.93 ?

r r rr 1.91: A and C are perpendicular, so A ? C = 0. A C + A C = 0, , which gives xx yy xy rr B ? C = 15.0, so – 3.5C + 7.0C = 15.0 xy We have two equations in two unknowns C and C . Solving gives xy xy

1.92: rr A?ùB = AB sin ? rr A ?ù B (- 5.00)2 + (2.00)2 sin ? = = = 0.5984 AB (3.00)(3.00) = sin-1(0.5984) = 36.8o ?

1.93: a) Using Equations (1.21) and (1.27), and recognizing that the vectors rr r A, B, and C do not have the same meanings as they do in those equations,

( r r ) r ( )i^ ( A B – A B ) ^ + (A B )k^ ) r A?ùB?C=AB-AB+ j -AB?C yz zy zx xz xy yx yzx zyx zxy xzy xyz yxz A similar calculation shows that r (r r) A?B?ùC=ABC-ABC+ABC-ABC+ABC-ABC xyz xzy yzx yxz zxy zyx and a comparison of the expressions shows that they are the same.

b) Although the above expression could be used, the form given allows for ready rr () compuation of A ?ù B the magnitude is AB sin ? = 20.00 sin 37.0o and the direction is, from the right-hand rule, in the +z-direction, and so ( r r ) r (20.00) (6.00) A ?ù B ? C = + sin 37.0o = +72.2.

(L + l) (W + w) = LW + lW + Lw, (L ÔÇô l) (W ÔÇô w) = LW ÔÇô lW ÔÇô Lw, where the common terms wl have been omitted. The area and its uncertainty are then WL ?? (lW + Lw), so the uncertainty in the area is a = lW + Lw.

b) The fractional uncertainty in the area is a lW + Wl l w = =+ A WL L W , the sum of the fractional uncertainties in the length and width.

c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional uncertainty in the volume is v lWH + LwH + LWh l w h = =++, V LW H L W H

1.95: The receiver’s position is (+ 1.0 + 9.0 – 6.0 + 12.0)^ + (- 5.0 + 11.0 + 4.0 + 18.0) ^ = (16.0)^ + (28.0) ^. i jij

The vector from the quarterback to the receiver is the receiver’s position minus the quarterback’s position, or (16.0)^ + (35.0) ^ , a vector with magnitude ij (16.0)2 + (35.0)2 = 38.5, given as being in yards. The angle is arctan(16.0 ) = 24.6o to the 35.0 right of downfield.

ii) In AU, (1.3087)2 + (-.4423)2 + (-.0414)2 = 1.3820 iii) In AU,

c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Equations (1-18) and (1.21), ? (-0.3182)(1.3087 – 0.3182) + (-0.9329)(-0.4423 – 0.9329) + ? = arccos ? ? (0.9857)(1.695)

d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90o.

The law of cosines (see Problem 1.88) gives the distance as (138 )2 + (77 )2 + 2(138 )(77 ) o = ly ly ly ly cos154.4 76.2 ly,

where the supplement of 25.6o has been used for the angle between the direction b) Although the law of cosines could be used again, it’s far more convenient to use the law of sines (Appendix B), and the angle is given by ? sin 25.6o ? arcsin? 138ly?=51.5o,180o-51.5o=129o, ? 76.2 ly ? where the appropriate angle in the second quadrant is used.

r 1.98: Define S = Ai^ + B^j + Ck^ rr r?S=(xi^+y^+zk^)?(Ai^+B^j+Ck^) j = Ax + By + Cz rr r If the points satisfy Ax + By + Cz = 0, then r ? S = 0 and all points r are r perpendicular to S .

1.00?ù103m-63m,andsothemagnitudeoftheaveragevelocityis 1.00 ?ù103 m – 63 m 4.75 s 1.00?ù10 m = 169 m s 3 b) 5.90 s

2.2: a) The magnitude of the average velocity on the return flight is

(5150 ?ù 103 m) (13.5 da) (86, 400 s da) b) Because the bird ends up at the starting point, the average velocity for the round trip is 0.

2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be ? 105 km hr ? ? 70 km hr ?

2.4: The eastward run takes (200 m 5.0 m s) = 40.0 s and the westward run takes (280 m 4.0 m s) = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = 4.4 m s to two significant figures. b) The net displacement is 80 m west, so the average velocity is (80 m 110.0 s) = 0.73 m s in the ÔÇôx-direction (-i^).

2.5: In time t the fast runner has traveled 200 m farther than the slow runner: Slow runner has run (5.50 m s)t = 1570 m.

t = t + 33s sp d d = vt ? t = v dd = + 33s vv sp dd = + 33s 3.5 km 6.5 km ss d = 250 km

2.7: a) The van will travel 480 m for the first 60 s and 1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of 14.0 m s . b) The first stage of the journey takes 240 m = 30 s and the second stage of the journey takes 8.0 m s (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s. c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same.

2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 5.60m-0= 20.8m-0= 20.8m-5.60m= 2.80 m s b) 5.2 m s c) 7.6 m s 2.00 s 4.00 s 2.00 s

2.9: a) At t = 0, x = 0 , so Eq (2.2) gives 11 x (2.4 m s 2 )(10.0 s) 2 – (0.120 m s 3 )(10.0 s) 3 av t (10.0 s) 2

b) From Eq. (2.3), the instantaneous velocity as a function of time is

= – 2= 2 – 3 2 v 2bt 3ct (4.80 m s )t (0.360 m s )t , x

so i) v (0) = 0, x ii) v (5.0 s) = (4.80 m s )(5.0 s) – (0.360 m s )(5.0 s) 2 = 15.0 m s , 23 x

x c)Thecarisatrestwhenv=0.Therefore(4.80ms2)t-(0.360ms3)t2=0.The x

This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward (positive slope and positive velocity), but becoming less so.

2.11: Time (s) 0 2 4 6 8 10 12 14 16 Acceleration (m/s2) 0 1 2 2 3 1.5 1.5 0

a) The acceleration is not constant, but is approximately constant between the times t = 4 s and t = 8 s.

2.12: The cruising speed of the car is 60 km hr = 16.7 m s . a) = 1.7 m s (to 16.7 m s 2 10 s two significant figures). b) 0-16.7 m s = -1.7 m s 2 c) No change in speed, so the acceleration 10 s is zero. d) The final speed is the same as the initial speed, so the average acceleration is zero.

((7.0 m s)t ) k^ r r (t ) = -(5.0 m s)ti^ + (10.0 m s)t^j + – (3.0 m s 2 )t 2

The velocity vector is the time derivative of the displacement vector:

r d r (t ) = (-5.0 m s)i^ + (10.0 m s) ^j + (7.0 m s – 2(3.0 m s 2 )t ) ^ k dt and the acceleration vector is the time derivative of the velocity vector:

r d 2 r (t ) = -6.0 m s2 k^ dt 2 At t = 5.0 s: ((7.0m s)(5.0s) ))k^ r r (t ) = -(5.0 m s)(5.0 s)i^ + (10.0 m s)(5.0 s) ^j + – (-3.0 m s 2 )(25.0 s 2 = (-25.0 m)i^ + (50.0 m) ^ – (40.0 m) ^ jk r d r (t ) = (-5.0 m s)i^ + (10.0 m s) ^j + ((7.0 m s – (6.0 m s 2 )(5.0 s)) k^ dt = (-5.0 m s)i^ + (10.0 m s) ^j – (23.0 m s) ^ k r d 2 r (t ) = -6.0 m s 2 k^ dt 2 (b) The velocity in both the x- and the y-directions is constant and nonzero; thus (c) The object’s acceleration is constant, since t does not appear in the acceleration vector.

dx 2.15:v==2.00cms-(0.125cms2)t x dt dv a = x = -0.125 cm s2 x dt xx x

c) Set x = 50.0 cm and solve for t. This gives t = 0 and t = 32.0s. The turtle returns d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm. x

x Set x = 40.0 cm and solve for t : t = 36.4 s (other root to the quadratic equation is negative x

e) 2.16: Use of Eq. (2.5), with t = 10 s in all cases, a) ((5.0m/s)-(15.0m/s))/(10s)=-1.0m/s2 b) ((-15.0m/s)-(-5.0m/s))/(10s)=-1.0m/s2 c) ((-15.0m/s)-(-15.0m/s))/(10s)=-3.0m/s2.

2.17: a) Assuming the car comes to rest from 65 mph (29 m/s) in 4 seconds, x

b) Since the car is coming to a stop, the acceleration is in the direction opposite to the velocity. If the velocity is in the positive direction, the acceleration is negative; if the velocity is in the negative direction, the acceleration is positive.

(3.00 m s ) + (0.100 m s3 ) (0) = 3.00 m s , and the velocity at t = 5.00 s is (3.00 m s ) + (0.100 m s3 ) (5.00 s)2 = 5.50 m s , so Eq. (2.4) gives the average acceleration as (5.50 m s) – (3.00 m s) 2 (5.00 s)

b) The instantaneous acceleration is obtained by using Eq. (2.5),

dx v = = (9.60 m s )t – (0.600 m s )t 5 26 x dt dv 26 x dt There are two times at which v = 0 (three if negative times are considered), given by t = 0 and t4 = 16 s4. At t = 0, x = 2.17 m and ax = 9.60 m s 2. When t4 = 16 s4, x=(2.17m)+(4.80ms2) (16s4)ÔÇô(0.100)ms6)(16s4)3/2=14.97m, b)

2.21: a) Equating Equations (2.9) and (2.10) and solving for v0, 2(x – x ) 2(70 m) 0x x t 7.00 s b) The above result for v0x may be used to find v – v 15.0 m s – 5.00 m s a = x 0x = =1.43 m s , 2 x t 7.00 s or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to eliminate v0x and solving for ax, ? v x – x ? ? 15.0 m s 70.0 m ? 2

0x v 2 ((173 mi hr) ( )) 2 0.4470 m s ax=2(x-x)= 2((307ft)(1m)) =32.0ms, x 1 mi hr 2 0 3.281ft

b) The time can be found from the above acceleration, v (173 mi hr) (0.4470 m s ) ax 32.0 m s2 The intermediate calculation may be avoided by using Eq. (2.14), again with v0x = 0, 2(x-x ) 2((307ft( 1m )) v (173 mi hr) 0.4470 m s x 1 mi hr

v = 0, a = v02 < a . Taking x = 0, FromEq.(2.13),with x x 2(x-x0) max 0 2.23:

v2 ((105kmhr)(1ms)(3.6kmhr))2 2a 2(250 m s 2 ) max

2.24: In Eq. (2.14), with x ÔÇô x0 being the length of the runway, and v0x = 0 (the plane x t 8s

2.25: a) From Eq. (2.13), with v = 0, 0x v2 (20 m s)2 x 2(x – x ) 2(120 m) 0

0 0x x x-x=vt+1at2givesa=6.67fts2 0 0x 2 x x

slowing down: 0 0x x x x 0x x 0 x

2 b) 1320 6.67 ft s , v 0 0x x x x 0x x 0 x x

At 4.0 s, v = +2.7 cm s (to the right) At 7.0 s, v = -1.3 cm s (to the left) b)a=slopeofv-tgraph=-8.0cm/s=-1.3cms2whichisconstant 6.0 s c) x = area under v-t graph

First 4.5 s: x=A +A Rectangle Triangle )?? ? 1 (4.5 s)?? 6 cm ?? = = (4.5 s 2 cm ? + 22.5 cm ?s?2 ?s?

From 0 to 7.5 s: 1 ( )? cm ? 1 ( )? cm ? d= 6s?8 ?+ 1.5s?2 ?=25.5cm 2?s?2?s?

9s,theaccelerationisconstant(fromthegraph)andequalto45ms-20ms=6.3ms2.From 4s t = 9 s to t = 13 s the acceleration is constant and equal to 2 4s b) In the first five seconds, the area under the graph is the area of the rectangle, (20 m)(5 s) = 100 m. Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s + 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is 230 m. Between t = 9 s and t = 13 s, the area under the triangle is (1 2)(45 m s)(4 s) = 90 m , and so the total distance in the first 13 s is 320 m.

2.32: 2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time the speed will be

b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel (1 2)(18 km s)(900 s) = 8100 km , so the distance traveled at non-constant speed is 16,200 km and the fraction of the distance traveled at constant speed is 16,200 km 1- = 0.958, 384,000 km c)Thetimespentatconstantspeedis384,000km-16,200km=2.04?ù104sandthetimespent 18 km s during both the period of acceleration and deceleration is 900 s, so the total time required forthetripis2.22?ù104s,about6.2hr.

1 (1.60ms2)(14.0s)2=156.8m 2 (from Eq. (2.12), with x0 = 0, v0x = 0), and has attained a speed of (1.60 m s2)(14.0s)=22.4 m s.

During the 70-second period when the train moves with constant speed, the train travels (22.4 m s) (70 s) = 1568 m. The distance traveled during deceleration is given by Eq. (2.13), with v = 0,v = 22.4 m s and a = -3.50 m s2 , so the train moves a distance x 0x x x – x = -(22.4 m / s)2 = 71.68 m. The total distance covered in then 156.8 m + 1568 m + 71.7 m 0 2(-3.50 m/s2 ) = 1.8 km.

In terms of the initial acceleration a1, the initial acceleration time t1, the time t2 during which the train moves at constant speed and the magnitude a2 of the final acceleration, the total distance xT is given by 1 1 (a t ) 2 ? a t ?? a t ? x = a t 2 + (a t )t + 1 1 = ? 1 1 ?? t + 2t + 1 1 ?, 2 2 | a2 | ? 2 ?? 2 T 11 11 2 1 | a2 | ? which yields the same result.

c) d) From Fig. (2.35), the graphs have the same slope at t = 2 s . e) Car A passes car B when they have the same position and the slope of curve A is greater than that of curve B in Fig. (2.30); this is at t = 3 s. f) Car B passes car A when they have the same position and the slope of curve B is greater than that of curve A; this is at t = 1 s.

truck’s constant speed, and the car’s position is given by xC = (1/2) aCt2. Equating the two expressions and dividing by a factor of t (this reflects the fact that the car and the truck are at the same place at t = 0) and solving for t yields 2v 2(20.0 m s) t = T = = 12.5 s aC 3.20 m s2 and at this time xT = xC = 250 m.

b) aCt = (3.20 m/s2)(12.5 s) = 40.0 m/s (See Exercise 2.37 for a discussion of why the car’s speed at this time is twice the truck’s speed.) c)

The car and the motorcycle have gone the same distance during the same time, so their average speeds are the same. The car’s average speed is its constant speed vC, and for constant acceleration from rest, the motorcycle’s speed is always twice its average, or 2vC. b) From the above, the motorcyle’s speed will be vC after half the time needed to catch the car. For motion from rest with constant acceleration, the distance traveled is proportional to the square of the time, so for half the time one-fourth of the total distance has been covered, or d 4.

2.38: a) An initial height of 200 m gives a speed of 60 m s when rounded to one significant figure. This is approximately 200 km/hr or approximately 150 mi hr . (Different values of the approximate height will give different answers; the above may be interpreted as slightly better than order of magnitude answers.) b) Personal experience will vary, but speeds on the order of one or two meters per second are reasonable. c) Air resistance may certainly not be neglected.

2.39: a) From Eq. (2.13), with v = 0 and a = -g , yy v = 2g(y-y)= 2(9.80ms2)(0.440m)=2.94ms, 0y 0 which is probably too precise for the speed of a flea; rounding down, the speed is about b) The time the flea is rising is the above speed divided by g, and the total time is twice this; symbolically,

2g(y-y) 2(y-y) 2(0.440m) = 0= 0= = t 2 2 2 0.599 s, g g (9.80m/s2) or about 0.60 s.

2.40: Using Eq. (2.13), with downward velocities and accelerations being positive, v 2 = y

(0.8 m s )2 + 2(1.6 m s 2 )(5.0 m) = 16.64 m 2 s 2 (keeping extra significant figures), so vy = 4.1 m s .

d=(12)gt2,sot=2dg.Ifdismeasuredincentimeters,thereactiontimeis 22- g 980 cm s2

v+v2-2g(y-y) t = 0y 0y 0 g

(5.00ms)+(5.00ms)-2(9.80ms2)(-12.0m) 2 = (9.80 m s 2 ) = 2.156 s, keeping an extra significant figure. The average velocity is then 12.0 m = 5.57 m s , down. 2.156 s As an alternative to using the quadratic formula, the speed of the ring when it hits the groundmaybeobtainedfromv2=v2-2g(y-y),andtheaveragevelocityfound y 0y 0 v +v from y 0 y ; this is algebraically identical to the result obtained by the quadratic formula. 2

b) While the ring is in free fall, the average acceleration is the constant acceleration due togravity,9.80m/s2down.

1 c) y = y + v t – gt 2 0 0y 2 1 0=12.0m+(5.00ms)t- (9.8ms2)t2 2 Solve this quadratic as in part a) to obtain t = 2.156 s.

d)v2=v2-2g(y-y)=(5.00ms)2-2(9.8ms2)(-12.0m) y 0y 0

i) at t = 0.250 s, y = (40.0 m) + (5.00 m s )(0.250 s) ÔÇô (1/2)(9.80 m s 2 )(0.250 s)2 = 40.9 m, 2 vy = (5.00 m s ) ÔÇô (9.80 m s )(0.250 s) = 2.55 m s and ii) at t = 1.00 s, y = (40.0 m) + (5.00 m/s)(1.00 s) ÔÇô (1/2)(9.80 m/s2)(1.00 s)2 = 40.1 m, 2 vy = (5.00 m s ) ÔÇô (9.80 m s )(1.00 s) = ÔÇô 4.80 m s .

b) Using the result derived in Example 2.8, the time is (5.00 m s) + (5.00 m s)2 – 2(9.80 m s2)(0 – 40.0 m) t = = 3.41 s. 2 (9.80 m s )

c) Either using the above time in Eq. (2.8) or avoiding the intermediate calculation by using Eq. (2.13), v2=v2-2g(y-y)=(5.00ms)2-2(9.80ms2)(-40.0m)=809m2s2, y 0y 0 vy = 28.4 m s .

y 0y 11 b)y=vt-gt2=(-6.00ms)(2.00s)-(9.80ms2)(2.00s)2=-31.6m,withthe 0y 22 c) v2=v2-2g(y-y)=(6.00ms)2-2(9.80ms2)(-10.0m)=232m2s2,sov=15.2m y 0y 0 y

2.46: a) The vertical distance from the initial position is given by

1 0y 2 solving for v0y, y 1 (-50.0 m) 1 2 0y t 2 (5.00 s) 2

b) The above result could be used in v2 = v2 – 2g(y – y ), with v = 0, to solve for y y 0y 0 ÔÇô y0 = 10.7 m (this requires retention of two extra significant figures in the calculation 2 for v0y). c) 0 d) 9.8 m s , down.

2.47: a) b) c) The most direct way to 9.80 m s 2 ave d) (283 m s) (1.40 s) = but 40 g = 392 m s , so the figures are not consistent. 22 202 m s

2 2.48: a) From Eq. (2.8), solving for t gives (40.0 m s ÔÇô 20.0 m s )/9.80 m s = 2.04 s. b) Again from Eq. (2.8), 40.0 m s – (-20.0 m s) 2 9.80 m s

c) The displacement will be zero when the ball has returned to its original vertical position, with velocity opposite to the original velocity. From Eq. (2.8), 40 m s – (-40 m s) 2 9.80 m s

(This ignores the t = 0 solution.) 2 d) Again from Eq. (2.8), (40 m s )/(9.80 m s ) = 4.08 s. This is, of course, half the time found in part (c).

the magnitude of g, and with g one-tenth as large, the height would be ten times higher, or 7.5 m. b) Similarly, if the ball is thrown with the same upward speed, it would go ten times as high, or 180 m. c) The maximum height is determined by the speed when hitting the ground; if this speed is to be the same, the maximum height would be ten times as large, or 20 m.

v = v + ?t ?t dt 21 t1 ? = v + (t 2 – t 2 ) 11 2 ?? = v – t2 + t2 11 22 = (5.0 m s ) ÔÇô (0.6 m s3 )(1.0 s)2 + (0.6 m s3 ) t2 = (4.40 m s ) + (0.6 m s3 ) t2.

At t2 = 2.0 s, the velocity is v2 = (4.40 m s ) + (0.6 m s3 )(2.0 s)2 = 6.80 m/s, or 6.8 m s b) From Eq. (2.16), the position x2 as a function of time is x = x + ? t v dt 2 1 t1 x =(6.0m)+?t((4.40m/s)+(0.6m/s3)t2)dt t1 3 (0.6 m s ) 11 3 At t = 2.0 s, and with t1 = 1.0 s,

x = (6.0 m) + (4.40 m s )((2.0 s) ÔÇô (1.0 s)) + (0.20 m s3 )((2.0 s)3 ÔÇô (1.0 s)3) = 11.8 m.

?t A B v = ( At – Bt 2 ) dt = t 2 – t 2 = (0.75 m s 3 )t 3 – (0.040 m s 4 )t 3 x 0 23 ?t ? A B ? A B x = ? t 2 – t 3 ? dt = t 3 – t 4 = (0.25 m s 3 )t 3 – (0.010 m s 4 )t 4 . 0 ? 2 3 ? 6 12

b) For the velocity to be a maximum, the acceleration must be zero; this occurs at t=0 and t = A = 12.5 s . At t=0 the velocity is a minimum, and at t=12.5 s the velocity is B

b) h = Area underv – tgraph max ?A +A Triangle Rectangle 1 ? cm ? ? (1.3 ms)?133 ? + (2.5 ms – 1.3 ms)(133 cm s) 2 ? s? ? 0.25 cm

c) a = slope of vÔÇô t graph 133 cm s 2 a(0.5ms)?a(1.0ms)? =1.0?ù105cms 1.3 ms a (1.5 ms) = 0 because the slope is zero.

d) h = area under vÔÇô t graph 1 ? cm ? -3 h(0.5ms)?A = (0.5ms)?33 ?=8.3?ù10 cm Triangle 2 ? s ? ? = 1 = ?ù -2 h (1.0 ms) A (1.0 ms)(100 cm s) 5.0 10 cm Triangle 2

vertical lines at t = 2.5s and t = 7.5s. This area is 1 (4.00 cm s2 + 8.00 cm s2 )(7.5 s – 2.5 s) = 30.0 cm s 2

2.54: a) To average 4 mi hr , the total time for the twenty-mile ride must be five hours, so the second ten miles must be covered in 3.75 hours, for an average of 2.7 mi hr . b) To average 12 mi hr , the second ten miles must be covered in 25 minutes and the average speed must be 24 mi hr . c) After the first hour, only ten of the twenty miles have been covered, and 16 mi hr is not possible as the average speed.

The velocity and acceleration of the particle as functions of time are v(t)=(9.00ms3)t2-(20.00ms2)t+(9.00ms) x

x b) The particle is at rest when the velocity is zero; setting v = 0 in the above expression and using the quadratic formula to solve for the time t,

(20.0 m s3 ) ?? (20.0 m s3 )2 – 4(9.0 m s3 )(9.0 m s) t= 2(9.0 m s3 )

and the times are 0.63 s and 1.60 s. c) The acceleration is negative at the earlier time and positive at the later time. d) The velocity is instantaneously not changing when the acceleration is zero; solving the above expression for a (t) = 0 gives x

20.00 m s2 18.00 m s3 Note that this time is the numerical average of the times found in part (c). e) The greatest distance is the position of the particle when the velocity is zero and the acceleration is negative; this occurs at 0.63 s, and at that time the particle is at (3.00ms3)(0.63s)3-(10.0ms2)(0.63s)2+(9.00ms)(0.63s)=2.45m.

(In this case, retaining extra significant figures in evaluating the roots of the quadratic equation does not change the answer in the third place.) f) The acceleration is negative at t = 0 and is increasing, so the particle is speeding up at the greatest rate at t = 2.00 s and slowing down at the greatest rate at t = 0. This is a situation where the extreme values of da a function (in the case the acceleration) occur not at times when = 0 but at the dt endpoints of the given range.

20.0 s 15 s c) Her net displacement is zero, so the average velocity has zero magnitude. d)50.0m=1.43ms.Notethattheanswertopart(d)istheharmonicmean,notthe 35.0 s arithmetic mean, of the answers to parts (a) and (b). (See Exercise 2.5).

2.57: Denote the times, speeds and lengths of the two parts of the trip as t1 and t2, v1 and v2, and l1 and l2.

a) The average speed for the whole trip is l + l l + l (76 km) + (34 km) 1 2 = 1 2 = ( 76 km ) + ( 34 km ) = 82 km h, t + t (l v ) + (l v ) 1 2 1 1 2 2 88 km h 72 km h

b) Assuming nearly straight-line motion (a common feature of Nebraska highways), the total distance traveled is l1ÔÇôl2 and

v ave l – l (76 km) – (34 km) t +t 1 2 88 km h 72 km h

( 31.4 km hr to three significant figures.) 2.58: a) The space per vehicle is the speed divided by the frequency with which the cars 96 km h 2400 vehicles h

An average vehicle is given to be 4.5 m long, so the average spacing is 40.0 m ÔÇô 4.6 m = 35.4 m.

constant speed (5.1 s) as t2. The constant speed is then at1, where a is the unknown acceleration. The total l is then given in terms of a, t1 and t2 by

1 l = at 2 + at t , 1 12 2 and solving for a gives l (100 m) a = = = 3.5 m s2 . (12)t2+tt (12)(4.0s)2+(4.0s)(5.1s) 1 12

2.60: a) Simple subtraction and division gives average speeds during the 2-second b) The average speed increased by 1.6 m s during each 2-second interval, so the 2 acceleration is 0.8 m s .

c) From Eq. (2.13), with v0 = 0, v = 2(0.8 m s )(14.4 m) = 4.8 m s. Or, 2

recognizing that for constant acceleration the average speed of 5.6 m/s is the speed one 2 second after passing the 14.4-m mark, 5.6 m s ÔÇô (0.8 m s )(1.0 s) = 4.8 m s . d) With both the acceleration and the speed at the 14.4-m known, either Eq. (2.8) e) From Eq. (2.12), x ÔÇô x0 = (4.8 m s )(1.0 s) + 1 (0.8 m s 2 )(1.0 s)2 = 5.2 m. This 2

is also the average velocity (1/2)(5.6 m s + 4.8 m s ) times the time interval of 1.0 s.

2.61: If the driver steps on the gas, the car will travel If the brake is applied, the car will travel (20ms)(3.0s)+(12)(-3.8ms2)(3.0s)2=42.9m, so the driver should apply the brake.

m ?3651d??24h??3600s? 2.62: a) d=ct=(3.0?ù108 )(1y)? 4 ?? ?? ? s ? 1y ?? 1d ?? 1h ? = 9.5?ù1015 m

b) d=ct=(3.0?ù108m)(10-9s)=0.30m s d 1.5?ù1011 m c) t = = = 500 s = 8.33 min c 3.0 ?ù108 m s d 2(3.84 ?ù108 m) d) t = = = 2.6 s c 3.0?ù108 m s d 3?ù109 mi e) t = = = 16,100 s = 4.5 h c 186,000 mi s

be found from: v2=v2+2a(x-x)=0+2(1.6ms2)(30m)=96m2s2 A 0A A 0

v = 96 m2 s2 = 9.80 m s A A’s time to cover the first 30 m is thus: v – v 9.80 m s t = A 0A = = 6.13 s aA 1.6 m s2

and A’s total time for the race is: (350 – 30) m 6.13 s + = 38.8 s 9.80 m s B’s speed at the end of 30 m is found from:

v2=v2+2a(x-x)=0+2(2.0ms2)(30m)=120m2s2 B 0B B 0

v=120m2s2=10.95ms B B’s time for the first 30 m is thus v – v 10.95 m s t = B 0B = = 5.48 s aB 2.0 m s2 and B’s total time for the race is: (350 – 30) m 5.48s + = 34.7 s 10.95 m s

0x x 0x x x x This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: 0x x 0 x-x=vt+1at2gives150m=(25s2)a+(12.5s2)aanda=4.0ms2 0 0x 2 x x x x

moves with constant velocity) is stationary. Then, the passenger train has an initial relative velocity of vrel,0 = 10 m s . This relative speed would be decreased to zero after the relative separation had decreased to vr2el,0 = + 500 m. Since this is larger in 2 arel magnitude than the original relative separation of 200 m, there will be a collision. b) The time at which the relative separation goes to zero (i.e., the collision time) is found by solving a quadratic (see Problems 2.35 & 2.36 or Example 2.8). The time is given by t = 1 (v – v2 + 2ax ) rel,0 rel,0 rel,0 a =(10s2m)(10ms-100m2s2-40m2s2) = (100 s)((1 – 0.6 ).

Substitution of this time into Eq. (2.12), with x0 = 0, yields 538 m as the distance the passenger train moves before the collision.

2.67: The total distance you cover is 1.20 m + 0.90 m = 2.10 m and the time available is 1.20 m = 0.80 s . Solving Eq. (2.12) for ax, 1.50 m s ( x – x ) – v t (2.10 m) – (0.80 m s)(0.80 s) a = 2 0 0x = 2 = 4.56 m s . 2

20 m s 2.5 m s 2 At this time, the officer is v 2 (20 m s) 2 1 2a 2(2.5 m s2)

This could also be found from (1/ 2)a1t12 , where t1 is the time found for the acceleration. At this time the car has moved (15 m s )(8.0 s) = 120 m, so the officer is 40 m behind the car.

a) The remaining distance to be covered is 300 m ÔÇô x1 and the average speed is (1/2)(v1 + v2) = 17.5 m s , so the time needed to slow down is 360 m – 80 m = 16.0 s, 17.5 m s and the total time is 24.0 s.

c) The officer slows from 20 m s to 15 m s in 16.0 s (the time found in part (a)), so 2 the acceleration is ÔÇô0.31 m s .

TTT t = 2(40.0 m) = 6.17 s (2.10 m s) b) The car has moved a distance 1 a 3.40 m s2 a t 2 = C x = 40.0 m = 64.8 m, 2 C a 1 2.10 m s2 T

1 120 2 The cars collide when x1 = x2; setting the expressions equal yields a quadratic in t,

1 at 2 + v t – D = 0, 0 2 the solutions to which are t = v2 + 2aD – v , t = – v2 + 2aD – v 0000 aa The second of these times is negative and does not represent the physical situation.

b) v = at = ( v 2 + 2aD – v ) 100

1 spreader when the brakes are applied, and the magnitude of the acceleration will be a=v12.Travellingat25ms,Juanisx=37m-(25ms)(0.80s)=17mfromthe 2 x1 2 spreader, and the speed of the car (and Juan) at the collision is obtained from

? v 2 ? ? x ? ? 17 m ? v 2 = v 2 – 2a x = v 2 – 2? 1 ? x = v 2 – v 2 ? 2 ? = (25 m s) 2 – (20 m s) 2 ? ? x 0x x 2 0x 2 0x 1 ? x1 ? ? ? ? ? 2 x 21 m 1

= 301 m2 s2 x b) The time is the reaction time plus the magnitude of the change in speed (v -v) 0

divided by the magnitude of the acceleration, or v – v 25 m s – 17.4 m s t = t + 2 0 x = (0.80 s) + 2 (21 m) = 1.60 s. flash reaction v 2 1 (20 m s) 2 0

2.72: a) There are many ways to find the result using extensive algebra, but the most straightforward way is to note that between the time the truck first passes the police car and the time the police car catches up to the truck, both the truck and the car have travelled the same distance in the same time, and hence have the same average velocity over that time. Since the truck had initial speed 3 v and the average speed is vp, the 2p 2p

3 3 b) The velocity is zero at t = ?? ? (a = 0 at t = 0, but this is an inflection point, not an ?

extreme). The extreme values of x are then ? ? ? ?3 ? 2 ?3 ? ? 3 ?3? 3 ? ?? The positive value is then 1

2 ? (4.00 m s)3 ? 2 2 ? 3 3? 3 ? 2.00 m s ?

2.75: a) The particle’s velocity and position as functions of time are

v (t ) = v + ?0t ((-2.00 m s 2 ) + (3.00 3 )t ) dt ms x 0x

? 3? ? 3.00 m s ?t 2 , = 0 x (2.00 m s ? 2 ?? v- 2+ )t ? x(t ) = ?0t v (t )dt = v t – (1.00 m s 2 )t 2 + (0.50 m s 3 )t 3 x 0x

= t (v – (1.00 m s 2 )t + (0.50 m s 3 )t 2 ), 0x where x0 has been set to 0. Then, x(0) = 0, and to have x(4 s) = 0, v – (1.00 m s 2 )(4.00 s) + (0.50 m s 3 )(4.00s) 2 = 0, 0x

0x x 2.76: The time needed for the egg to fall is

2 h 2(46.0 m – 1.80 m) t = = = 3.00 s, 9 (9.80 m s2)

return. The total time is T = t + t = 2.5 s . Let y be the height of the building, then, 12 y = 1 gt 2 and y = v t There are three equations and three unknowns. Eliminate t , solve 2 1 s 2. 2 for t , and use the result to find y. A quadratic results: 1 gt 2 + v t – v T = 0. If 1 2 1 s1 s

2a Here, t = t , a = 1 2 g = 4.9 m s 2 , b = v = 340 m s , and c = -v T = -(340 m s)(2.5 s) = -850 m 1ss

Then upon substituting these values into the quadratic formula,

-(340ms)?? (340ms)2-4(4.9ms2)(-850m) t= 12 2(4.9 m s ) t1 = ( 2(4.9 m/s2 ) ) = 2.42 s – 340 m s )??(363.7 m s . The other solution, ÔÇô71.8 s has no real physical meaning.

Then,y=1gt2=1(9.8ms2)(2.42s)2=28.6m.Check:(28.6m)(340ms)=.08s,the 212 time for the sound to return.

2.78: The elevators to the observation deck of the Sears Tower in Chicago move from the ground floor to the 103rd floor observation deck in about 70 s. Estimating a single (103)(3.5m) = floor to be about 3.5 m (11.5 ft), the average speed of the elevator is 5.15 m s. 70 s Estimating that the elevator must come to rest in the space of one floor, the acceleration 0 -(5.15 m s ) = – 2 22 2 3.5 m

b) The required speed would be v=v2+2g(y-y)=(25ms)2+2(9.80ms2)(-21.3m)=14.4ms, 00

window is h ave 0 0 t

The distance l from the roof to the top of the window is then v2 ((1.90 m) /(0.420 s) – (1 2)(9.80 m s2 )(0.420 s))2 l = 0 = = 0.310 m. 2g 2(9.80 m s2 ) An alternative but more complicated algebraic method is to note that t is the difference between the times taken to fall the heights l + h and h, so that

2(l + h) 2l gg Squaring the second expression allows cancelation of the l terms,

(1 2)gt 2 + 2 gt 2l 2 = h, which is solved for 1 ? h ?2 l = ? – (1 2) gt ? , 2g ? t ? which is the same as the previous expression.

2 g = 2((9.80 m s)2 ) 2.81: a) The football will go an additional v 5.00 m s = 1.27 m above the window, so 2

2.83: a) From Eq. (2.14), with v0=0, yy0 b) The height above the release point is also found from Eq. (2.14), with v =7.59ms,v =0anda =-g, 0y y y

v2 (7.59 m s)2 h = = = 2.94 m 0y 2g 2(9.80ms2)

(Note that this is also (64.0 cm) ( ).The height above the ground is then 5.14 m. 2 45 m s g

c) See Problems 2.46 & 2.48 or Example 2.8: The shot moves a total distance 2.20 m ÔÇô1.83 m = 0.37 m, and the time is

11 y= gt2= (9.8ms2)(9.0s2)=44.1m 22 To reach her ears after 3.0 s, the sound must therefore have traveled a total distance of h + (h – 44.1) m = 2h – 44.1 m ,where h is the height of the cliff. Given 340 m/s for the speed of sound: 2h – 44.1 m = (340 m s)(3.0 s) = 1020 m , which gives b)Wecanusev2=v2+2g(y-y)tofindtheteacher’sfinalvelocity.Thisgives y 0y 0 yy

reaches an upward velocity of v=v+at=(5.0ms2)(10.0s)=50.0ms y 0y y andaheightofy=1at2=1(5.0ms2)(10.0s)2=250matthemomenttheengineisshut 2y 2 off. To find the helicopter’s maximum height use v 2 = v 2 + 2a ( y – y ) y 0y y 0 Taking y = 250 m , where the engine shut off, and since v 2 = 0 at the maximum height: 0y -v2 y -y = 0y max 0 2g (50.0 m s)2 y = 250 m – = 378 m max 2(-9.8 m s2 ) or 380 m to the given precision.

b) The time for the helicopter to crash from the height of 250 m where Powers stepped out and the engine shut off can be found from: 1 y=y+vt+at2=250m+(50.0ms)t+1(-9.8ms2)t2=0 0 0y 2 y 2 where we now take the ground as y = 0 . The quadratic formula gives solutions of t = 3.67 s and 13.88 s, of which the first is physically impossible in this situation. Powers’ position 7.0 seconds after the engine shutoff is given by:

1 y=250m+(50.0ms)(7.0s)+ (-9.8ms2)(49.0s2)=359.9m 2 at which time his velocity is v = v + gt = 50.0 m s + (-9.80 m s )(7.0 s) = -18.6 m s 2 y 0y Powers thus has 13.88 – 7.0 = 6.88s more time to fall before the helicopter crashes, at his constant downward acceleration of 2.0m s2 . His position at crash time is thus: 1 y = y + v t + a t2 0 0y y 2 1 =359.9m+(-18.6ms)(6.88s)+ (-2.0ms2)(6.88s)2 2 = 184.6 m or 180 m to the given precision.

Last 1.0 s of fall: y-y=vt+1at2givesh4=v(1.0s)+(4.9ms2)(1.0s)2 0 0y 2 y 0y Motion from roof to y – y = 3h 4 : 0

0y y v2=v2+2a(y-y)givesv=2(9.80ms2)(3h4)=3.834hms y 0y y 0 y This is vy for the last 1.0 s of fall. Using this in the equation for the first 1.0 s gives h 4 = 3.834 h + 4.9 Let h = u 2 and solve for u : u = 16.5. Then h = u 2 = 270 m.

2.88: a) t + t = 10.0 s fall sound return t + t = 10.0 s (1) fs

d =d Rock Sound 1 gt 2 = v t f ss 2 1 (9.8 m s 2 )t 2 = (330 m s)t (2) fs 2

Combine(1)and(2):t=8.84s,t=1.16s fs m h = v t = (330 )(1.16 s) = 383 m ss s b) You would think that the rock fell for 10 s, not 8.84 s, so you would have thought it fell farther. Therefore your answer would be an overestimate of the cliff’s height.

0 y 0y y – y = v t + 1 a t 2 gives v = 11.31 m s 0 0y 2 y 0y Use this v0y in v = v + a t to solve for v : v = -20.5 m s y 0y y y y

b) Find the maximum height of the can, above the point where it falls from the scaffolding: 2 y 0y y 0 v 2 = v 2 + 2a ( y – y ) gives y – y = 6.53 m y 0y y 0 0 The can will pass the location of the other painter. Yes, he gets a chance.

falling for a time t0 before Superman’s leap (in this case, t0 = 5 s). Then, the height h of the building is related to t and t0 in two different ways: 1 – h = v t – gt 2 0y 2 = – 1 g (t + t )2 , 0 2

where v0y is Superman’s initial velocity. Solving the second t gives t g – t0. = 2h hg Solving the first for v0y gives v0y = – + t, and substitution of numerical values gives t2 t = 1.06 s and v0y = ÔÇô165 m s , with the minus sign indicating a downward initial velocity.

speed of the second part of the fall (with the Rocketeer supplying the upward acceleration), and assuming the student is a rest both at the top of the tower and at the v2 v2 ground, the distances fallen during the first and second parts of the fall are 1 and 1 , 2g 10g where v1 is the student’s speed when the Rocketeer catches him. The distance fallen in free fall is then five times the distance from the ground when caught, and so the distance above the ground when caught is one-sixth of the height of the tower, or 92.2 m. b) The student falls a distance 5H 6 in time t = 5H 3g , and the Rocketeer falls the same distance in time tÔÇôt0, where t0=5.00 s (assigning three significant figures to t0 is more or less arbitrary). Then,

5H 1 = v (t – t ) + g (t – t ) 2 , or 00 0 62 5H 6 1 (t – t ) 2 00 0

At this point, there is no great advantage in expressing t in terms of H and g algebraically; 00

? ? =?- 2 A takes the early lead. b) The cars are both at the origin at t = 0. The non-trivial solution is found by setting xA = xB, cancelling the common factor of t, and solving the quadratic for

] 2 2? Substitution of numerical values gives 2.27 s, 5.73 s. The use of the term ÔÇ£starting pointÔÇØ can be taken to mean that negative times are to be neglected. c) Setting vA = vB leads to a different quadratic, the positive solution to which is

2 6? d) Taking the second derivative of xA and xB and setting them equal, yields, 2? = 2? – 6?t . Solving, t = 2.67 s .

position be x0. Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student x1 and the bus x2 as functions of time are then 10 20

a) Setting x = x and solving for the times t gives 12 1 (v ) t= ?? v2-2ax 000 a ((5.0 m s) )(40.0 m) ) 1 = ?? (5.0 m s)2 – 2(0.170 m s2 (0.170 m s2 ) = 9.55 s, 49.3 s.

The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance 0

2= c) The results can be verified by noting that the x lines for the student and the bus intersect at two points:

d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus’s velocity is (0.170 m s2 )(49.3 s) = 8.38 m s.

quantities will be denoted symbolically. That is, 1 1( ) let y = h + v t – gt 2 , y = h – g t – t . In this case, t = 1.00 s . Setting 2 10200 22 y = y = 0, expanding the binomial (t – t )2 and eliminating the common term 12 0 2 0020 1 gt 2 t 1 gt – v 2 1 – 0 v 0 0 gt 0

Substitution of this into the expression for y and setting y = 0 and solving for h as a 11 function of v0 yields, after some algebra, ? 1 ?2 ? gt – v ? 0

2 0 (gt – v )2 00 a) Using the given value t = 1.00 s and g = 9.80 m s2 , 0

(4.9 )?? 4.9 m s – v ?2 ? 9.8 m s – v ? 0 This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see part (c)). The physical solution involves taking the negative 0

b) The above expression gives for i), 0.411 m and for ii) 1.15 km. c) As v0 approaches 9.8 m s , the height h becomes infinite, corresponding to a relative velocity at the time the second ball is thrown that approaches zero. If v > 9.8 m s , the first ball can never 0

catch the second ball. d) As v0 approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer (on its way down) to the top of the roof when the second ball is released. If v > 4.9 m s , the first ball will already have 0

passed the roof on the way down before the second ball is released, and the second ball can never catch up.

equations h = 1 gt2, 2 h = 1 g(t – t)2 can be solved for t. Eliminating h and taking the 232

square root, t = 3 , and t = t , and substitution into h = 1 gt 2 gives h = 246 m. t – t 2 1- 2 3 2 This method avoids use of the quadratic formula; the quadratic formula is a generalization of the method of ÔÇ£completing the squareÔÇØ, and in the above form, 32

b) The above method assumed that t >0 when the square root was taken. The negative root (with t = 0) gives an answer of 2.51 m, clearly not a ÔÇ£cliffÔÇØ. This would correspond to an object that was initially near the bottom of this ÔÇ£cliffÔÇØ being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer.

(5.3m)-(1.1m) v = = 1.4 m s , x,ave (3.0 s) (-0.5 m) – (3.4 m) y ,ave (3.0 s)

b)v=(1.4ms)2+(-1.3ms)2=1.91ms,or1.9mstotwosignificantfigures, ave 1.4

3.2: a) x = (v ) t = (-3.8 m s)(12.0 s) = -45.6 m and x,ave r = y,ave b)r=x2+y2=(-45.6m)2+(58.8m)2=74.4m.

vij 3.5: a) (-170 m s) – (90 m s) b) a = = -8.7 m s , 2 x,ave (30.0 s) (40 m s) – (110 m s) 2 y ,ave (30.0 s)

-8.7 3.6: a) a = (0.45 m s ) cos 31.0?? = 0.39 m s , a = ) sin 31.0?? = 0.23 , 2222 (0.45 m s m s xy

b) r =?^-2?t^=(2.4ms)i^-[(2.4ms2)t]^j vij r 2 aj j r c) At t = 2.0 s , the velocity is = (2.4 m s) ^ – (4.8 m s) ^ ; the magnitude is vij (2.4ms)2+(-4.8ms)2=5.4ms,andthedirectionisarctan(-4.8)=-63??.The 2.4 acceleration is constant, with magnitude 2.4 m s 2 in the – y -direction. d) The velocity vector has a component parallel to the acceleration, so the bird is speeding up. The bird is turning toward the – y -direction, which would be to the bird’s right (taking the + z – direction to be vertical).

0y 0 b)vt=0.385mc)v=v=1.10ms,v=-gt=-3.43ms,v=3.60ms,72.2?? x x 0x y below the horizontal.

g b) The bomb’s constant horizontal velocity will be that of the plane, so the bomb x

c) The bomb’s horizontal component of velocity is 60 m/s, and its vertical component d)

e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m above the bomb at impact.

v = 0, a = -9.80 m s , y – y = -h, t = 3.50 s 2 0y y 0 y – y = v t + 1 a t 2 gives h = 60.0 m 0 0y 2 y v = v sin 32.0??, y – y = -60.0 m, a = -9.80 m s = 2 0y 0 0 y y – y = v t + 1 a t 2 gives t = 3.55 s 0 0y 2 y 0x 0 x 0

3.12: Time to fall 9.00 m from rest: 1 y = gt 2 2 1 9.00m= (9.8ms2)t2 2 t=1.36s

Speed to travel 1.75 m horizontally: x=vt 0 1.75 m = v (1.36 s) 0

v = 1.3 m s 0 Use the vertical motion to find the time in the air: v = 0, a = -9.80 m s 2 , y – y = -(21.3 m – 1.8 m) = -19.5 m, t = ? 0y y 0 y – y = v t + 1 a t 2 gives t = 1.995 s 0 0y 2 y 0 x 0x 0 0x 2 x 0x b) v = 30.6 m s since a = 0 xx v = v + a t = -19.6 m s y 0y y

moment it leaves the tabletop and the time it will take for the ball to reach the floor (or rather, the rim of the cup). The latter can be determined simply by measuring the height of the tabletop above the rim of the cup and using y = 1 gt 2 to calculate the falling time. 2

The horizontal velocity can be determined (although with significant uncertainty) by timing the ball’s roll for a measured distance before it leaves the table, assuming that its speed doesn’t change much on the hard tabletop. The horizontal distance traveled while the ball is in flight will simply be horizontal velocity ?ù falling time. The cup should be placed at this distance (or a slightly shorter distance, to allow for the slowing of the ball on the tabletop and to make sure it clears the rim of the cup) from a point vertically below the edge of the table.

y 0y (15.0 m s) sin 45?? 2 9.80 m s b)UsingEquations(3.20)and(3.21)givesatt,(x,y)=(6.18m,4.52m): 1

23 c) Using Equations (3.22) and (3.23) gives at t , (v , v ) = (10.6 m s , 4.9 m s) : t , (10.6 m s , 0) t : (10.6 m s , – 4.9 m s), for 1xy 2 3 velocities, respectively, of 11.7 m s @ 24.8??, 10.6 m s @ 0?? and 11.7 m s @ -24.8??. Note that v is the same for all times, and that the y-component of velocity at t is x3 1

d) The parallel and perpendicular components of the acceleration are obtained from vrr vr v (a ? v )v v a ? v v v v || v 2 || v || Forprojectilemotion,v=-g^j, v?vr=-gv a so a , and the components of acceleration y

parallel and perpendicular to the velocity are t : -4.1 m s 2 , 8.9 m s 2 . t : 0, 9.8 m s 2 . 12 3

e) f) At t1, the projectile is moving upward but slowing down; at t2 the motion is instantaneously horizontal, but the vertical component of velocity is decreasing; at t3, the projectile is falling down and its speed is increasing. The horizontal component of velocity is constant.

0 b) Assuming a horizontal tabletop, v = 0 , and from Eq. (3.16), 0y 0x 0 c) On striking the floor, v = -gt = – 2gy = -3.83m s , and so the ball has a velocity y0 of magnitude 5.24 m s , directed 46.9?? below the horizontal.

d) Although not asked for in the problem, this y vs. x graph shows the trajectory of the tennis ball as viewed from the side.

3.17: The range of a projectile is given in Example 3.11, R = v2 sin 2? g . 00 a)(120ms)2sin110??(9.80ms2)=1.38km.b)(120ms)2sin110??(1.6ms2= ) 8.4 km .

g 9.80 m s2 2 2 2 y0 2g c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s d) v is constant at 20.0 m s , so (20.0 m s)(3.27 s) = 65.3 m . x

e) 3.19: a) v = (30.0 m s) sin 36.9?? = 18.0 m s ; solving Eq. (3.18) for t with y = 0 and 0y 0 y = 10.0 m gives (18.0ms)?? (18.0ms)2-2(9.80ms2)(10.0m) t = = 0.68 s, 2.99 s 2 9.80 m s b) The x-component of velocity will be (30.0 m s) cos 36.9?? = 24.0 m s at all times. The y-component, obtained from Eq. (3.17), is 11.3 m s at the earlier time and c) The magnitude is the same, 30.0 m s , but the direction is now 36.9?? below the horizontal.

b) The x-component of velocity is constant at v = (12.0 m s) cos 51.0?? = 7.55 m s . The x

y-component is v = (12.0 m s) sin 51.0?? = 9.32 m s at release and 0y

0y 0x f) 3.21: a) The time the quarter is in the air is the horizontal distance divided by the horizontal component of velocity. Using this time in Eq. (3.18), x gx 2 y-y =v – 0 0 y v 2v 2 0x 0x

gx 2 = tan? x – 0 2 2? v 2 cos 00

dart ? gd ? ? 2v2 cos2 ? dart 0 0 0?

Using the given values for d and ? to express this as a function of v , 00 ? 26.62 m 2 s 2 ? ? 2? ? v0 ? Then, a) y = 2.14 m , b) y = 1.45 m , c) y = -2.29 m . In the last case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal 3.23: a) With v = 0 in Eq. (3.17), solving for t and substituting into Eq. (3.18) gives y

v v 2 sin ? (30.0 m/s) sin 33.0?? 2 2 22 ( y – y ) = 0 y = 0 0 = = 13.6 m 0 2g 2g 2(9.80 m/s2 )

b) Rather than solving a quadratic, the above height may be used to find the time the rock takes to fall from its greatest height to the ground, and hence the vertical component ofvelocity,v=2yg=2(28.6m)(9.80m/s2)=23.7m/s,andsothespeedofthe y

v cos?t = 45.0 m 0 45.0 m cos? = = 0.600 (25.0 m/s)(3.00 s)

? = 53.1?? b) v = (25.0 m/s) cos 53.1?? = 15.0 m/s x v =0 y v = 15.0 m/s a = 9.80 m/s2 downward c) Find y when t = 3.00 s y=vsin?t-1gt2 0 2 1 = (25.0 m/s)(sin53.1??)(3.00s) – (9.80 m/s 2 )(3.00 s) 2 2 = 15.9 m v = 15.0 m/s = constant x

v = v sin ? – gt = (25.0 m/s)(sin 53.1??) – (9.80 m/s 2 )(3.00 s) = -9.41 y0

0y y 0 y – y = v t + 1 a t 2 gives y – y = 296 m 0 0y 2 y 0

b) In 6.00 s the balloon travels downward a distance y – y = (20.0 s)(6.00 s) = 120 m . 0

c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the rock and the basket is 176 m, so the rock is (176 m)2 + (90 m)2 = 198 m from the basket when it hits the ground.

d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 Just before the rock hits the ground, its vertical component of velocity is v = v + a t = 20.0 s + (9.80 m/s2 )(6.00 s) = 78.8 m/s , downward, relative to the ground. y 0y y The basket is moving downward at 20.0 m/s, so relative to the basket the rock has e) horizontal: 15.0 m/s; vertical: 78.8 m/s

3.26: a) horizontal motion: x – x = v t so t = 60.0 m 0 0 x (v0 cos 43??)t vertical motion (take + y to be upward): y-y=vt+1at2gives25.0m=(vsin43.0??)t+1(-9.80m/s2)t2 0 0y 2 y 0 2 Solving these two simultaneous equations for v and t gives v = 3.26 m/s and t = 2.51s . 00 b) v when shell reaches cliff: y

v = v + a t = (32.6 m/s) sin 43.0?? – (9.80 m/s 2 )(2.51 s) = -2.4 m/s y 0y y The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff.

Use the vertical motion to find the time it takes the suitcase to reach the ground: 0y 0 y 0 y – y = v t + 1 a t 2 gives t = 9.60 s 0 0y 2 y

square of the frequency, and hence inversely proportional to the square of the rotational period; tripling the centripetal acceleration involves decreasing the period by a factor of 3 , so that the new period T ? is given in terms of the previous period T by T ? = T / 3 .

3.29: Using the given values in Eq. (3.30), 4?2(6.38?ù106m) rad ((24 h)(3600 s/h)) 2 (Using the time for the siderial day instead of the solar day will give an answer that differs in the third place.) b) Solving Eq. (3.30) for the period T with a = g , rad

4?2(6.38?ù106m) 9.80 m/s2 3.30: 550 rev/min = 9.17 rev/s , corresponding to a period of 0.109 s. a) From Eq. (3.29), v = 2?R = 196 m/s . b) From either Eq. (3.30) or Eq. (3.31), T

rad 3.31: Solving Eq. (3.30) for T in terms of R and a , rad

3.32:a)UsingEq.(3.31),2?R=2.97?ù104m/s.b)EitherEq.(3.30)orEq.(3.31)gives T

rad 3.33: a) From Eq. (3.31), a = (7.00 m/s)2 /(15.0 m) = 3.50 m/s2 . The acceleration at the b) a = 3.50 m/s2 , the same as part (a), but is directed down, and still towards the center. c) From Eq. (3.29), T = 2?R / v = 2? (15.0 m)/(7.00 m/s) = 12.6 s .

rad tan b)

3.35: b) No. Only in a circle would a point to the center (See planetary motion in rad c) Where the car is farthest from the center of the ellipse.

3.36: Repeated use of Eq. (3.33) gives a) 5.0 = m/s to the right, b) 16.0 m/s to the left, and c) 13.0 = m/s to the left.

3.37: a) The speed relative to the ground is 1.5 m/s + 1.0 m/s = 2.5 m/s , and the time is 35.0 m/2.5 m/s = 14.0 s. b) The speed relative to the ground is 0.5 m/s, and the time is 70 s.

3.38: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream, so the total time the rower takes is 1.5 km 1.5 km 6.8 km/h 1.2 km/h 3.39: The velocity components are – 0.50 m/s + (0.40 m/s)/ 2 east and (0.40 m/s)/ 2 south, for a velocity relative to the earth of 0.36 m/s, 52.5?? south of west.

km/h, so the heading must be arcsin 80.8 = 14?? north of west. b) Using the angle found in 320 part(a),(320km/h)cos14??=310km/h.Equivalently, (320 km/h)2 – (80.0 km/h)2 = 310 km/h .

3.41: a) (2.0 m/s) 2 + (4.2 m/s) 2 = 4.7 m/s, arctan 2.0 = 25.5?? , south of east. 4.2 c)2.0m/s?ù190s=381m.

3.42: a) The speed relative to the water is still 4.2 m/s; the necessary heading of the boat is arcsin 2.0 = 28?? north of east. b) (4.2 m/s)2 – (2.0 m/s)2 = 3.7 m/s , east. d) 4.2 800 m/3.7 m/s = 217 s , rounded to three significant figures.

3.43: a) b) x : -(10 m/s) cos 45?? = -7.1 m/s. y := -(35 m/s) – (10 m/s)sin 45?? = -42.1 m/s . c) (-7.1 m/s) 2 + (-42.1 m/s) 2 = 42.7 m/s, arctan -42.1 = 80?? , south of west. -7.1

3.44: a) Using generalizations of Equations 2.17 and 2.18, v = v + ? t 3 , v = v + ?t – ? t 2 , and x = v t + ? t 4 , y = v t + t 2 – t 3 . b) Setting ?? x 0 x 3 y 0 y 2 0 x 12 0 y 2 6 v = 0 yields a quadratic in t, 0 = v + ?t – ? t 2 , which has as the positive solution y 0y 2 t = 1 [? + ? + 2v ? ]= 13.59 s, 2

?0 keeping an extra place in the intermediate calculation. Using this time in the expression for y(t) gives a maximum height of 341 m.

xy y b)Thespeedisv=(?2+4?2t2)1/2,dv/dt=0att=0.(Seepartdbelow.) c) r and v are perpendicular when their dot product is 0: (?t)(?)+(15.0m-?t2)?ù(-2?t)=?2t-(30.0m)?t+2?2t3=0.Solvethisfort: t=??(30.0m)(0.500m/s2)-(1.2m/s)2=+5.208s,and0s,atwhichtimesthestudentisat(6.25m, 22 2(0.500 m/s ) d) At t = 5.208 s , the student is 6.41 m from the origin, at an angle of 13?? from the x- axis.Aplotofd(t)=(x(t)2+y(t)2)1/2showstheminimumdistanceof6.41mat5.208 s:

e) In the x – y plane the student’s path is: r?? r 32

b) The positive time at which x = 0 is given by t 2 = 3? ? . At this time, the y-coordinate is ? 3?? 3(2.4 m/s)(4.0 m/s2 ) y = t 2 = = = 9.0 m 2 2? 2(1.6 m/s3) .

v2 ((88 km/h)(1m/s)/(3.6 km/h))2 a = = = 0.996 m/s2 ? 1 m/s2 2x 2(300 m) b) arctan( 15 m ) = 5.4?? . c) The vertical component of the velocity is 460 m-300 m (88km/h)(1m/s)15m=2.3m/s.d)Theaveragespeedforthefirst300mis44km/h,so 3.6 km/h 160 m the elapsed time is 300 m 160 m + = 31.1s, (44km/h)(1m/s)(3.6km/h) (88km/h)(1m/s)cos5.4??/(3.6km/h) or 31 s to two places.

a) The equations of motions are: 1 y = h + (v sin ?)t – gt 2 0 2 x = (v cos ?)t 0

v = v sin ? – gt y0 v = v cos ? x0 Note that the angle of 36.9o results in sin 36.9?? = 3/5 and cos 36.9?? = 4/5 . b) At the top of the trajectory, v = 0 . Solve this for t and use in the equation for y to y

= v sin ? = + (v sin ? ) – (v sin ? )2 find the maximum height: t 0 . Then, y h (v sin ?) 0 1 g 0 , which g 0g2g

=+22 = v sin ? ? = reduces to y h 0 . Using v 25gh / 8 , and sin 3 / 5 , this becomes 2g 0

y=h+(25gh/8)(3/5)2=h+9h,ory=25h.Note:Thisanswerassumesthaty=h.Taking 2 g 16 16 0 0 16 c) The total time of flight can be found from the y equation by setting y = 0 , assuming y = h , solving the quadratic for t and inserting the total flight time in the x equation to 0

find the range. The quadratic is 1 gt 2 – 3 v – h = 0 . Using the quadratic formula gives 2 50 ?? – 2 – 1 – ?? 9 ?À 25 gh + 16 gh = (3/5)v0 ( (3/5)v0) 4( g)( h) = = (3/5) 25gh/8 t 2 . Substituting v 25gh / 8 gives t 25 8 8 . 2( 1 g ) 0 g ( ) (3 ) 2

Collecting terms gives t: t = 1 9h ?? 25h = 1 h ?? 5 h . Only the positive root is 2 2g 2g 2 2g 2g

2g 0 8 5 2g

ground) and rearranging gives 2v 2 sin ? cos ? 2v 2 x2 – 0 0 0 x – 0x h = 0, gg The easier thing to do here is to recognize that this can be put in the form 2v0xv0y 2v2 x2 – x – 0x h = 0, gg the solution to which is v[ ] 0y 0y g x

3.53: The distance is the horizontal speed times the time of free fall, 2 y 2(90 m) x g (9.80 m/s2 )

0 0 x-y change in height. Substitution of numerical values gives v = 42.8 m/s . b) Using the 0

above algebraic expression for v in Eq. (3.27) gives 0 ? x ?2 y = x – ? ? (188.9 m) ? 188 m ? Using x = 116 m gives y = 44.1 m above the initial height, or 45.0 m above the ground, which is 42.0 m above the fence.

3.56:Theequationsofmotionsarey=(vsin?)t-1/2gt2andx=(vcos?)t,assuming 00 the match starts out at x = 0 and y = 0 . When the match goes in the wastebasket for the minimum velocity, y = 2D and x = 6D . When the match goes in the wastebasket for the maximum velocity, y = 2D and x = 7D . In both cases, sin ? = cos ? = 2 / 2. . To reach the minimum distance: 6D = 2 v t , and 2D = 2 v t – 1 gt 2 . Solving the first 20 202 equation for t gives t = 6D 2 . Substituting this into the second equation gives v0

2v 00 0

To reach the maximum distance: 7D = 2 v t , and 2D = 2 v t – 1 gt 2 . Solving the first 20 202 equation for t gives t = 7D 2 . Substituting this into the second equation v0

= – ( )2 gives2D7D1g7D2.Solvingthisforvgivesv=49gD/5=3.13gD,which, 2 v0 0 0

3.57: The range for a projectile that lands at the same height from which it was launched is R = v02 sin 2? , and the maximum height that it reaches is H = v02 sin ? . We must find R 2

g 2g

H= = sin?,D=6gDsin2? when D and v 6gD . Solving the height equation for , or 0 2g

a) Solving for v gives 0 gx2 / 2 cos2 ? x tan ? – y 0 0 0 b) Eliminating t between Equations 3.20 and 3.23 gives v as a function of x , y

v cos ? y00 00 Using the given values yields v = v cos ? = 8.28 m/s, v = -6.98 m/s, so x00 y

x, the solution to which is v2 cos ? ? 2gh ? x = 0 0 ?tan2? + ? g ? v02 cos ?0 ? 0 v cos ? [ ] 0000 g If h = 0 , the square root reduces to v sin ? , and x = R . b) The expression for x 00

becomesx=(10.2m)cos?+[sin2?+sin2?+0.98] 000 The angle ? = 90?? corresponds to the projectile being launched straight up, and there 0

is no horizontal motion. If ? = 0 , the projectile moves horizontally until it has fallen the 0

c) The maximum range occurs for an angle less than 45?? , and in this case the angle is about 36?? .

0 b) c) Using (14. m – 1.9 m) instead of h in the above calculation gives x = 6.3 m , so the man will not be hit.

3.61: a) The expression for the range, as derived in Example 3.10, involves the sine of twice the launch angle, and sin (2(90?? – ? )) = sin (180?? – 2? ) = sin 180?? cos 2? – cos 180?? sin 2? = sin 2? , 00000 and so the range is the same. As an alternative, using sin(90?? – ? ) = cos? and 0

cos(90??-?)=sin?intheexpressionfortherangethatinvolvestheproductofthesine 00 0

0 b) Again, the algebra is the same as that used in Problem 3.58; v = 8.4 m/s , at an angle x

A graph of y(t) vs. x(t) shows the trajectory of Mary Belle as viewed from the side:

d) In this situation it’s convenient to use Eq. (3.27), which becomes y=(1.327)x-(0.071115m-1)x2.Useofthequadraticformulagivesx=23.8m.

3.63: a) The algebra is the same as that for Problem 3.58, gx2 2cos2? (xtan? – y) 0 00 In this case, the value for y is – 15.0 m , the change in height. Substitution of numerical values gives 17.8 m/s. b) 28.4 m from the near bank (i.e., in the water!).

v2=v2cos2?+(vsin?-gt)2 0000 =v2(sin2? +cos2? )-2v sin? gt+(gt)2 00000

y 0y y 0 v2 = v2 + 2a ( y – y ) gives y – y = 81.6 m y 0y y 0 0 b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in c) Use the vertical motion of the rocket to find the time it is in the air. 0y y y v = v + a t gives t = 8.164 s y 0y y 0 0x d) Relative to the ground the rocket has initial velocity components v = 30.0 m/s and 0x 0y e) (i)

Relative to the cart, the rocket travels straight up and then straight down (ii)

Time the ball is in the air: x (b) v (runner) = v (ball) xx 6.0 m/s = (20.0 m/s) cos? cos ? = 0.300 ? = 72.5??

y = v sin ?t – 1 gt 2 02 1 – 45.0 m = (20.0 m/s)(sin72.5??)t – (9.80 m/s 2 )t 2 2

a) Use the vertical motion of the boulder to find the time it takes it to fall 20 m to the 0y y 0 y – y = v t + 1 a t 2 gives t = 2.02 s 0 0y 2 y The rock must travel 100 m horizontally during this time, so x – x 100 m 0x t 2.20 s b) The rock travels downward 45 m in going from the cliff to the plain. Use this vertical motion to find the time: 0y y 0 y – y = v t + 1 a t 2 gives t = 3.03 s 0 0y 2 y During this time the rock travels horizontally x-x =v t=(49m/s)(3.03s)=150m 0 0x The rock lands 50 m past the foot of the dam.

3.68: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the bagels to fall 43.9 m from rest. Get the time to fall: 1 y = gt 2 2 1 43.9m= (9.80m/s2)t2 2 t = 2.99 s

So she has been jogging for 9.00 s + 2.99 s = 12.0 s . During this time she has gone x = vt = (3.05 m/s)(12.0 s) = 36.6 m . Bruce must throw the bagels so they travel 36.6 m horizontally in 2.99 s x = vt 36.6 m = v(2.99 s) v = 12.2 m/s (b) 36.6 m from the building.

a) The vertical motion of the shell is unaffected by the horizontal motion of the tank. Use the vertical motion of the shell to find the time the shell is in the air: 0y 0 y 0 y – y = v t + 1 a t 2 gives t = 8.86 s 0 0y 2 y Relative to tank #1 the shell has a constant horizontal velocity v cos? = 246.2 m/s . 0

Relative to the ground the horizontal velocity component is 246.2 m/s + 15.0 m/s = 261.2 m/s . Relative to tank #2 the shell has horizontal velocity component 261.2 m/s – 35.0 m/s = 226.2 m/s . The distance between the tanks when the shell was fired is the (226.2 m/s)(8.86 s) = 2000 m that the shell travels relative to tank b) The tanks are initially 2000 m apart. In 8.86 s tank #1 travels 133 m and tank #2 travels 310 m, in the same direction. Therefore, their separation increases by 310m-183m=177m.So,theseparationbecomes2180m(roundingto3significant figures).

3.70: The firecracker’s falling time can be found from the usual 2h t= g The firecracker’s horizontal position at any time t (taking the student’s position as x = 0 ) is x = vt – 1 at 2 = 0 when cracker hits the ground, from which we can find that t = 2v / a . 2

2 0 2g 2g maximum height above the floor is 2.23 m. b) Use of the result of Problem 3.59 gives 3.84 m. c) The algebra is the same as that for Problems 3.58 and 3.62. The distance y is 3.05m-1.83m=1.22m,and (9.80 m/s 2 )(4.21 m) 2 0 2cos235??((4.21m)tan35??-1.22m) d) As in part (a), but with the larger speed, The distance from the basket is the distance from the foul line to the basket, minus half the range, or Note that an extra figure in the intermediate calculation was kept to avoid roundoff error.

3.72: The initial y-component of the velocity is flight is t = 2 y / g . The initial x-component is the initial velocity is then x2g v = 2gy + 0 2y v0 x = v = 2gy , and the time the pebble is in 0y

4-5 triangle, and thus, a = (3 / 5) g and a = (4 / 5) g during the “boost” phase of the flight. xy Hence this portion of the flight is a straight line at an angle of 53.1?? to the horizontal. After time T, the rocket is in free flight, the acceleration is a = 0 and a = g , and the xy familiar equations of projectile motion apply. During this coasting phase of the flight, the trajectory is the familiar parabola.

b)Duringtheboostphase,thevelocitiesare:v=(3/5)gtandv=(4/5)gt,both xy straightlines.Aftert=T,thevelocitiesarev=(3/5)gT,ahorizontalline,and x

v =(4/5)gT-g(t-T),anegativelyslopinglinewhichcrossestheaxisatthetimeof y

c) To find the maximum height of the rocket, set v = 0 , and solve for t, where t = 0 y

the amount the enemy has pulled away in that time. Symbolically, R = x + v t = x + v R , where v is the velocity of the enemy relative to the hero, t 0 E/H 0 E/H v E/H 0x

is the time of flight, v is the (constant) x-component of the grenade’s velocity, as 0x measured by the hero, and R is the range of the grenade, also as measured by the hero. Using Eq. (3-29) for R, with sin 2? = 1 and v = v / 2 , 0 0x 0 v2 v 0 E/H 0 E/H 0 0 gg This quadratic is solved for 1 v = ( 2v + 2v2 + 4gx ) = 61.1km/h, 0 E/H E/H 0 2 where the units for g and x have been properly converted. Relative to the earth, the x- 0

componentofvelocityis90.0km/h+(61.1km/h)cos45??=133.2km/h,they-component, thesameinbothframes,is(61.1km/h)sin45??=43.2km/h,andthemagnitudeofthe velocity is then 140 km/h.

3.75:a) x2+y2=(Rcos?t)2+(Rsin?t)2=R2(cos2?t+sin2?t)=R2, b) v=-?Rsin?t,v=?Rcos?t, xy and so the dot product rr r ? v = xv + yv xy =(Rcos?t)(-?Rsin?t)+(Rsin?t)(?Rcos?t) =?R(-cos?tsin?t+sin?tcos?t) = 0.

c) a =-?2Rcos?t=-?2x, a =?2Rsin?t=-?2y, xy rr d)v2=v2+v2=(-?Rsin?t)2+(?Rcos?t)2=?2R2(sin2?t+cos2?t)=?2R2,andso xy v = ?R .

dv d = v2 + v2 xy dt dt (1/2)d(v2+v2) = dt x y v2 + v2 xy

va+va v2 + v2 xy b) Using the numbers from Example 3.1 and 3.2, dv (-1.0m/s)(-0.50m/s2)+(1.3m/s)(0.30m/s2) = = 0.54 m/s. dt (-1.0 m/s)2 + (1.3 m/s)2 The acceleration is due to changing both the magnitude and direction of the velocity. If the direction of the velocity is changing, the magnitude of the acceleration is larger than vv the rate of change of speed. c) v ? a = v a + v a , v = v 2 + v 2 , and so the above form xx yy x y vv dt

b) To find the velocity components, take the derivative of x and y with respect to time: v=R?(1-cos?t),andv=R?sin?t.Tofindtheaccelerationcomponents,takethe xy xy x y c) The particle is at rest (v = v = 0) every period, namely at t = 0, 2? / ?, 4? / ?,…. At yx that time, x = 0, 2?R, 4?R,..; and y = 0. The acceleration is a = R? 2 in the 2 2 1/ 2 222

3.78: A direct way to find the angle is to consider the velocity relative to the air and the velocity relative to the ground as forming two sides of an isosceles triangle. The wind direction relative to north is half of the included angle, or arcsin(10 / 50) = 11.53??, east of north.

is possible, and even entertaining, but hardly necessary. The relative speed of the brothers is 70 km/h, and as they are initially 42 km apart, they will reach each other in six-tenths of an hour, during which time the pigeon flies 30 km.

3.80: a) The drops are given as falling vertically, so their horizontal component of velocity with respect to the earth is zero. With respect to the train, their horizontal component of velocity is 12.0 m/s, west (as the train is moving eastward). b) The verticalcomponent,ineitherframe,is(12.0m/s)/(tan30??)=20.8m/s,andthisisthe magnitude of the velocity in the frame of the earth. The magnitude of the velocity in the frame of the train is (12.0 m/s) 2 + ( 20.8 m/s) 2 = 24 m/s. This is, of course, the same as (12.0 m/s) / sin 30??.

3.81: a) With no wind, the plane would be 110 km west of the starting point; the wind has blown the plane 10 km west and 20 km south in half an hour, so the wind velocity is (20 km/h)2 + (40 km/h)2 = 44.7 km/h at a direction of arctan(40 / 20) = 63?? south of west. b) arcsin(40 / 220) = 10.5?? north of west.

3.83:a)Thepositionoftheboltis3.00m+(2.50m/s)t-1/2(9.80m/s2)t2,andthe positionoftheflooris(2.50m/s)t.Equatingthetwo,3.00m=(4.90m/s2)t2.Therefore t=0.782s. b)Thevelocityoftheboltis2.50m/s-(9.80m/s2)(0.782s)=-5.17m/s relative to Earth, therefore, relative to an observer in the elevator v = -5.17 m/s – 2.50 m/s = -7.67 m/s. c) As calculated in part (b), the speed relative to Earth is 5.17 m/s. d) Relative to Earth, the distance the bolt travelled is (2.50m/s)t-1/2(9.80m/s2)t2=(2.50m/s)(0.782s)-(4.90m/s2)(0.782s)2=-1.04m

= 5310 km = 3.84: Air speed of plane 804.5 km/h 6.60 h With wind from A to B: t + t = 6.70 h AB BA Same distance both ways: 5310 km (804.5 km/h + v )t = = 2655 km w AB 2 (804.5 km/h + v )t = 2655 km w BA Solve (1), (2), and (3) to obtain wind speed v : w

r v Juan relative to the ground. This velocity is due north and has magnitude J/G, J/G r v the ball relative to the ground. This vector is 37.0?? east of north and has magnitude B/G, B/G r v the ball relative to Juan. We are asked to find the magnitude and direction of this B/J, rrrrrr The relative velocity addition equation is v = v + v so v = v – v . B/G B/J J/G, B/J B/G J/G v = +v sin 37.0?? = 7.222 m/s B/Jx B/G v =+v cos37.0??-v =1.584m/s B/Jy B/G J/G B/J

3.86:a)v = 2gh= 2(9.80m/s2)(4.90m)=9.80m/s. b)v /g=1.00s. c)The 0y 0y

speed relative to the man is (10.8 m/s)2 – (9.80 m/s)2 = 4.54 m/s, and the speed relative to the hoop is 13.6 m/s (rounding to three figures), and so the man must be 13.6 m in front of the hoop at release. d) Relative to the flat car, the ball is projected at an angle ?=tan-1( )=65??.Relativetothegroundtheangleis?=tan-1( )=35.7?? 9.80 m/s 9.80 m/s 4.54 m/s 4.54 m/s+9.10 m/s

? (80 m)2 c) The slower rise will tend to reduce the time in the air and hence reduce the radius. The slower horizontal velocity will also reduce the radius. The lower speed would tend to increase the time of descent, hence increasing the radius. As the bullets fall, the friction effect is smaller than when they were rising, and the overall effect is to decrease the radius.

3.88: Write an expression for the square of the distance (D2) from the origin to the particle, expressed as a function of time. Then take the derivative of D2 with respect to t, and solve for the value of t when this derivative is zero. If the discriminant is zero or negative,thedistanceDwillneverdecrease.Followingthisprocess,sin-1 8/9=70.5??.

0 equation describing the incline is y = x tan ?. Setting these equal and factoring out the x = 0 root (where the projectile is on the incline) gives a value for x ; the range 0

measured along the incline is ?2v2? ?cos2(?+?)? ? g ? ? cos ? ? b) Of the many ways to approach this problem, a convenient way is to use the same sort of “trick”, involving double angles, as was used to derive the expression for the range along a horizontal incline. Specifically, write the above in terms of ? = ? + ?, as ? 2v 2 ? ? g cos 2 ? ? The dependence on ? and hence ? is in the second term. Using the identities sin?cos?=(1/2)sin2?andcos2?=(1/2)(1+cos2?),thistermbecomes This will be a maximum when sin(2? – ?) is a maximum, at 2? – ? = 2? + ? = 90??, or ? = 45?? – ? / 2. Note that this reduces to the expected forms when ? = 0 (a flat incline, ? = 45?? and when ? = -90?? (a vertical cliff), when a horizontal launch gives the greatest distance).

? gx ? 1 ? 2v2 ? cos2 (? + ? ) 0 Denote the dimensionless quantity gx / 2v2 by ?; in this case 0

(9.80m/s2)(60.0m)cos30.0?? 2(32.0 m/s) 2 The above relation can then be written, on multiplying both sides by the product cos? cos(? + ? ), ? cos? sin?cos(?+?)=sin(?+?)cos?- , cos(? + ? ) and so ? cos? cos(? + ? )

Thetermontheleftissin((?+?)-?)=sin?,sotheresultofthiscombinationis sin?cos(?+?)=?cos?.

Although this can be done numerically (by iteration, trial-and-error, or other methods), 2

specifically, then 1 (sin(2? + ? ) + sin(-? )) = ? cos? , 2 with the net result that a) For ? = 30??, and ? as found above, ? = 19.3?? and the angle above the horizontal is ? + ? = 49.3??. For level ground, using ? = 0.2871, gives ? = 17.5??. b) For ? = -30??, the same ? as with ? = 30?? may be used (cos30?? = cos(-30??)), giving ? = 13.0?? and ? + ? = -17.0??.

R (see Figure 3.23). By considering the isosceles triangle formed by the two velocity vectors, the magnitude vr is seen to be 2v sin(? / 2), so that v v ? v t ? 10 m/s aave = 2 sin? ? = sin(1.0 / s ? t) t ? 2R ? t Usingthegivenvaluesgivesmagnitudesof9.59m/s2,9.98m/s2and10.0m/s2.The instantaneous acceleration magnitude, v2 / R = (5.00 m/s)2 /(2.50 m) = 10. m/s2 is indeed approached in the limit at t ? 0. The changes in direction of the velocity vectors are given by ? = v t and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle R

The x-position of the plane is (236 m/s)t and the x-position of the rocket is (236m/s)t+1/2(3.00)(9.80m/s2)cos30??(t-T)2.Thegraphsofthesetwohavetheform,

If we take y = 0 to be the altitude of the airliner, then y(t)=-1/2gT2-gT(t-T)+1/2(3.00)(9.80m/s2)(sin30??)(t-T)2fortherocket.This graph looks like

By setting y = 0 for the rocket, we can solve for t in terms of T,0=-(4.90m/s2)T2-(9.80m/s2)T(t-T)+(7.35m/s2)(t-T)2.Usingthequadratic formula for the variable x = t – T, we find x = t – T = (9.80 m/s )T + (9.80 m/s T ) +(4)(7.35 m/s )(4.9)T 2 or t = 2 22 2 2.72 T . Now, using 2(7.35 m/s2 ) the condition that x – x = 1000 m, we find rocket plane Therefore T = 5.15 s.

3.93: a) Taking all units to be in km and h, we have three equations. We know that heading upstream v – v = 2 where v is the speed of the curve relative to water c/w w/G c/w

and v / is the speed of the water relative to the ground. We know that heading wG downstream for a time t, (v + v )t = 5. We also know that for the bottle c/w w/G v (t + 1) = 3. Solving these three equations for v = x, v = 2 + x, therefore w/G w/G c/w (2+x+x)t=5or(2+2x)t=5.Alsot=3/x-1,so(2+2x)(3-1)=5or x

be parallel, and the angle between them is zero. b) The forces form the sides of a right isosceles triangle, and the angle between them is 90?? . Alternatively, the law of cosines may be used as F2+F2=(2F)-2F2cos?, 2

from which cos? = 0 , and the forces are perpendicular. c) For the sum to have 0 magnitude, the forces must be antiparallel, and the angle between them is 180?? .

4.2: In the new coordinates, the 120-N force acts at an angle of 53?? from the – x -axis, or 233?? from the + x -axis, and the 50-N force acts at an angle of 323?? from the + x – axis.

a) The components of the net force are R = (120 N) cos 233?? + (50 N) cos 323?? = -32 N x

y b) R = R2 + R2 = 128 N, arctan(124 ) = 104?? . The results have the same magnitude, x y -32 and the angle has been changed by the amount (37??) that the coordinates have been rotated.

4.3: The horizontal component of the force is (10 N) cos 45?? = 7.1 N to the right and the vertical component is (10 N) sin 45?? = 7.1 N down.

Geometric: From the law of cosines, the magnitude of the resultant is The angle between the resultant and dog A’s rope (the angle opposite the side corresponding to the 250-N force in a vector diagram) is then ? sin 120??(300 N) ? ? (494 N) ? Components: Taking the +x -direction to be along dog A’s rope, the components of the resultant are R = (270 N) + (300 N) cos 60?? = 420 N x

R = (300 N)sin 60?? = 259.8 N, y 420 4.6: a) F + F = (9.00 N) cos 120?? + (6.00 N) cos (-126.9??) = -8.10 N 1x 2 x 1y 2 y

xy 4.10: a) The acceleration is a = 2x = 2(11.0m) = 0.88 m / s2 . The mass is then t2 (5.00s)2 a 0.88 m/s 2 b) The speed at the end of the first 5.00 seconds is at = 4.4 m/s , and the block on the frictionless surface will continue to move at this speed, so it will move another vt = 22.0m in the next 5.00 s.

(keeping an extra figure). At t = 2.00 s , the speed is at = 3.13m/s and the position is at 2 / 2 = vt / 2 = 3.13 m . b) The acceleration during this period is also 1.563 m/s2 , and the speedat7.00sis3.13m/s+(1.563m/s2)(2.00s)=6.26m/s.Thepositionatt=5.00sis x=3.13m+(3.13m/s)(5.00s-2.00s)=125m,andatt=7.00sis 12.5 m + (3.13 m/s)(2.00 s) + (1/2)(1.563 m/s2 )(2.00 s)2 = 21.89 m, or 21.9 m to three places.

x 0x 2 0x x x

r 4.13: a) ? F = 0 b), c), d) 4.14: a) With v = 0 , 0x v2 (3.00?ù106 m/s)2 x 2x 2(1.80?ù10-2 m) b)t=vx=3.00?ù106m/s=1.20?ù10-8s.Notethatthistimeisalsothedistancedividedby a x 2.50?ù1014 m / s 2 –

F F F ? 160 ? 4.16: a= = = g=? ?(9.80m/s2)=22.0m/s2. m w / g w ? 71.2 ?

4.17: a) m = w / g = (44.0 N) /(9.80 m/s2 ) = 4.49 kg b) The mass is the same, 4.49 kg, and the weight is (4.49 kg)(1.81 m/s2 ) = 8.13 N.

4.19: F = ma = (55 kg)(15 m / s2 ) = 825 N. The net forward force on the sprinter is exerted by the blocks. (The sprinter exerts a backward force on the blocks.)

4.20: a) the earth (gravity) b) 4 N, the book c) no d) 4 N, the earth, the book, up e) 4 N, the hand, the book, down f) second g) third h) no i) no j) yes k) yes l) one (gravity) m) no

4.21: a) When air resistance is not neglected, the net force on the bottle is the weight of the bottle plus the force of air resistance. b) The bottle exerts an upward force on the earth, and a downward force on the air.

4.22: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, ? F = 620 N-650 N = – 2 upward and also of magnitude 650 N. 2 0.452 m/s . The passenger’s m 650 N /9.80 m/s acceleration is 0.452 m / s2 , downward.

F (the force on m due to m ) and F (the force on m due to m ) form an AB A B BA B A (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them.

4.25: The ball must accelerate eastward with the same acceleration as the train. There must be an eastward component of the tension to provide this acceleration, so the ball hangs at an angle relative to the vertical. The net force on the ball is not zero.

For the truck: The box’s friction force on the truck bed and the truck bed’s friction force on the box form an action-reaction pair. There would also be some small air-resistance force action to the left, presumably negligible at this speed.

b) For the chair, a = 0 so ? F = ma gives y yy n – mg – F sin 37?? = 0 n = 142 N

T is the force exerted by the rope and f is the force the ground exerts on the tricycle. g

spot and the wagon T ? is the force exerted by the rope. T and T ? form a third-law action-reaction pair, rr T = -T ?.

vave (v0 / 2) 350 m / s b)F=ma=(1.80?ù10-3kg)(350m/s)=848N.(Usinga=v2/2xgivesthesame (7.43?ù10-4 s) 0 result.)

rr 4.31: Take the + x -direction to be along F and the + y -direction to be along R . Then 1 r F = -1300 N and F = 1300 N , so F = 1838 N , at an angle of 135?? from F . 2x 2y 2 1

withy-component(140N)sin30??-(100N)sin60??=-16.6N.Forthechildtoexertthe smallest possible force, that force will have no x-component, so the smallest possible force has magnitude 16.6 N and is at an angle of 270?? , or 90?? clockwise from the + x -direction.

= ? F = 100 N cos 60??+140 N cos 30?? = = = 2 = b) m 85.6 kg. w mg (85.6 kg)(9.80 m / s ) 840 N. . a 2.0 m / s2

4.34: The ship would go a distance v2 v2 mv2 (3.6 ?ù107 kg)(1.5 m / s)2 0 = 0 = 0 = = 506.25 m, 2a 2(F/m) 2F 2(8.0?ù104N) so the ship would hit the reef. The speed when the tanker hits the reef is also found from 2(8.0 ?ù104 N)(500 m) v = v2 – (2Fx / m) = (1.5 m/s)2 – = 0.17 m/s, 0 (3.6 ?ù107 kg) so the oil should be safe.

y 0y y 0 v = 0 at the maximum height, y – y = 1.2 m, a = -9.80 m/s2 , so y 0y 0y av c)

F – w = ma av av F =w+ma =890N+(890N/9.80m/s2)(16.2m/s2) av av F = 2.36?ù103 N av

F = F – mg (upward) net b) When the upward force has its maximum magnitude F (the breaking strength), max the net upward force will be F – mg and the upward acceleration will be max F – mg F 75.0 N m m 4.80 kg

4.38: a) w = mg = 539 N b) r Downward velocity is decreasing so a is upward and the net force should be upward. air

b) T=ma=(4.00kg)(2.50m/s2)=10.0N 1 c) r d) F – T = m a 2 F=T+ma=10.0N+(6.00kg)(2.50m/s2)=25.0N 2

4.40: a) The force the astronaut exerts on the rope and the force that the rope exerts on the astronaut are an action-reaction pair, so the rope exerts a force of 80.0 N on the astronaut. b) The cable is under tension. c) a = F = 80.0 N = 0.762 m / s 2 . d) There is no m 105.0 kg net force on the massless rope, so the force that the shuttle exerts on the rope must be 80.0 N (this is not an action-reaction pair). Thus, the force that the rope exerts on the m 9.05?ù104 kg

b) Differentiating, the velocity as a function of time is v(t)=(1.80?ù104m/s2)t-(2.40?ù105m/s3)t2,so v(0.025s)=(1.80?ù104m/s2)(0.025s)-(2.40?ù105m/s3)(0.025s)2 =3.0?ù102m/s.

c) The acceleration as a function of time is a(t)=1.80?ù104m/s2-(4.80?ù105m/s3)t,

so(i)att=0,a=1.8?ù104m/s2,and(ii)a(0.025s)=6.0?ù103m/s2,andtheforcesare (i) ma = 2.7 ?ù104 N and (ii) ma = 9.0 ?ù103 N.

b) speeding up: w > F and the net force is downward slowing down: w < F and the net force is upward c) Denote the y-component of the acceleration when the thrust is F by a and the y- 11 component of the acceleration when the thrust is F by a . The forces and accelerations 22 are then related by 1 12 2 Dividing the first of these by the second to eliminate the mass gives F-w a 1 = 1, F -w a 22 and solving for the weight w gives aF-aF a -a 12 In this form, it does not matter which thrust and acceleration are denoted by 1 and which by 2, and the acceleration due to gravity at the surface of Mercury need not be found. Substituting the given numbers, with + y upward, gives

(1.20m/s2)(10.0?ù103N)-(-0.80m/s2)(25.0?ù103N) w = = 16.0 ?ù103 N. 1.20m/s2-(-0.80m/s2)

In the above, note that the upward direction is taken to be positive, so that a is negative. 2

Also note that although a is known to two places, the sums in both numerator and 2

a) The engine is pulling four cars, and so the force that the engine exerts on the first car is 4m . b), c), d): Similarly, the forces the cars exert on the car behind are rrr 3m a , 2m a and – m a . e) The direction of the acceleration, and hence the direction of r a

4.44: a) If the gymnast climbs at a constant rate, there is no net force on the gymnast, so b) No motion is no acceleration, so the tension is again the gymnast’s weight. c) T – w = T – mg = ma = m r d) T – w = T – mg = ma = -m r the tension), so T = m( g – a ) .

r a (the acceleration is upward, the same direction as the r a (the acceleration is downward, the same opposite as

4.45: a) The maximum acceleration would occur when the tension in the cables is a maximum, F T – mg T 28,000 N a = net = = – g = – 9.80 m / s 2 = 2.93 m / s 2 . m m m 2200 kg

b) The acceleration while the knees are bending is v2 (7.80 m / s)2 2 y 2(0.60 m)

c) The net force that the feet exert on the ground is the force that the ground exerts on the feet (an action-reaction pair). This force is related to the weight and acceleration by F-w=F-mg=ma,soF=m(a+g)=(75.0kg)(50.6m/s2+9.80m/s2)=4532N.As a fraction of his weight, this force is F = (a + 1) = 6.16 (keeping an extra figure in the mg g

intermediate calculation of a). Note that this result is the same algebraically as (3.10 m + 1). 0.60 m

4.47: a) b) The acceleration of the hammer head will be the same as the nail, a=v2/2x=(3.2m/s)2/2(0.45cm)=1.138?ù103m/s2.Themassofthehammerheadis 0

itsweightdividedbyg,4.9N/9.80m/s2=0.50kg,andsothenetforceonthehammer headis(0.50kg)(1.138?ù103m/s2)=570N.Thisisthesumoftheforcesonthehammer head; the upward force that the nail exerts, the downward weight and the downward 15-N force. The force that the nail exerts is then 590 N, and this must be the magnitude of the force that the hammer head exerts on the nail. c) The distance the nail moves is .12 m, so the acceleration will be 4267 m / s2 , and the net force on the hammer head will be 2133 N. The magnitude of the force that the nail exerts on the hammer head, and hence the magnitude of the force that the hammer head exerts on the nail, is 2153 N, or about 2200 N.

a) The net force on a point of the cable at the top is zero; the tension in the cable must b) The net force on the cable must be zero; the difference between the tensions at the top and bottom must be equal to the weight w, and with the result of part (a), there is no c) The net force on the bottom half of the cable must be zero, and so the tension in the cable at the middle must be half the weight, w / 2 . Equivalently, the net force on the upper half of the cable must be zero. From part (a) the tension at the top is w, the weight of the top half is w / 2 and so the tension in the cable at the middle must be d) A graph of T vs. distance will be a negatively sloped line.

4.49: a) b) The net force on the system is 200 N – (15.00 kg)(9.80 m / s2 ) = 53.0 N (keeping three figures), and so the acceleration is (53.0 N) /(15.0 kg) = 3.53 m / s 2 , up. c) The net forceonthe6-kgblockis(6.00kg)(3.53m/s2)=21.2N,sothetensionisfoundfrom F – T – mg = 21.2 N , or T = (200 N) – (6.00 kg)(9.80 m / s2 ) – 21.2 N = 120 N . Equivalently, the tension at the top of the rope causes the upward acceleration of the rope andthebottomblock,soT-(9.00kg)g=(9.00kg)a,whichalsogivesT=120N.d) The same analysis of part (c) is applicable, but using 6.00 kg + 2.00 kg instead of the mass of the top block, or 7.00 kg instead of the mass of the bottom block. Either way gives T = 93.3 N .

The downward forces of magnitude 2ma and ma for the top and middle links are the reaction forces to the upward force needed to accelerate the links below. b) (i) The weight of each link is mg = (0.300 kg)(9.80 m / s2 ) = 2.94 N . Using the free- body diagram for the whole chain: F 12 N – 3(2.94 N) 3.18 N a = net = = = 3.53 m / s2 or 3.5 m / s2 3m 0.900 kg 0.900 kg (ii) The second link also accelerates at 3.53 m / s2 , so:

F = F – ma – 2mg = ma net top F =2ma+2mg=2(0.300kg)(3.53m/s2)+2(2.94N) top = 2.12 N + 5.88 N = 8.0 N

4.53: Differentiating twice, the acceleration of the helicopter as a function of time is r =(0.120m/s3)t^-(0.12m/s2)^, aik and at t = 5.0 s , the acceleration is r aik The force is then

[(0.60 )k^ ] r w r (2.75?ù105 N) F=ma= a= m/s2)i^-(0.12m/s2 g (9.80 m / s 2 ) =(1.7?ù104N)i^-(3.4?ù103N)k^.

function of time is a(t) = -6Bt , and so the Force as a function of time is F(t) = ma(t) = -6mBt .

dt as the unknown function v(t) is part of the integrand. The equation must be separated before integration; that is, C dv – dt = m v2 Ct 1 1 – =- + , mvv 0

where v is the constant of integration that gives v = v at t = 0 . Note that this form 00 shows that if v = 0 , there is no motion. This expression may be rewritten as 0 -1 dx ? 1 Ct ? v = = ? + ? , dt ? 0 ? vm which may be integrated to obtain m ? Ctv ? 0

0C?m? To obtain x as a function of v, the time t must be eliminated in favor of v; from the expression obtained after the first integration, 0 = 0 -1, so Ctv v mv m ?v ? 0 C ?v? b) By the chain rule, dv dv dv dv = = v, dt dx dt dx and using the given expression for the net force,

position. As the y-component of force is given as an explicit function of time, v and y y

mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope; for the pulley to be in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N.

5.2: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w. Two forces act on each mass: w down and T (= w) up.

5.3: a) The two sides of the rope each exert a force with vertical component T sin ? , and the sum of these components is the hero’s weight. Solving for the tension T, w (90.0 kg) (9.80 m s2 ) 2 sin ? 2 sin 10.0?? b) When the tension is at its maximum value, solving the above equation for the angle ? gives ? w ? ? (90.0 kg) (9.80 m s2 ? ? 2T ? ? 2(2.50 ?ù104 N) ?

5.4: The vertical component of the force due to the tension in each wire must be half of the weight, and this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical, so if the weight is w, w = 3w cos ? and ? = arccos 2 = 48??. 24 3

5.5: With the positive y-direction up and the positive x-direction to the right, the free- body diagram of Fig. 5.4(b) will have the forces labeled n and T resolved into x- and y- components, and setting the net force equal to zero, F=Tcos?-nsin?=0 x

y Solving the first for n = T cot ? and substituting into the second gives cos 2? ? cos 2 ? sin 2 ? ? T T +Tsin?=T? + ?= =w sin ? ? sin ? sin ? ? sin ?

?=W,orT=?=(4090kg)(9.8ms2)=?ù4 5.7: a) T cos W cos 5.23 10 N. B B cos 40?? AB

5.8:a)T=w,Tsin30??+Tsin45??=T=w,andTcos30??-Tcos45??=0.Since CABCAB sin45??=cos45??,addingthelasttwoequationsgivesT(cos30??+sin30??)=w,andso A

A 1.366 B A cos 45?? b) Similar to part (a), T = w, – T cos 60?? + T sin 45?? = w, and CAB T sin 60?? – T cos 45?? = 0. Again adding the last two, T = w = 2.73w, and A B A (sin60??-cos60??) B B cos 45??

5.10: The magnitude of the force must be equal to the component of the weight along the incline,orWsin?=(180kg)(9.80ms2)sin11.0??=337N.

12 12 5.12: If the rope makes an angle ? with the vertical, then sin? = 0.110 = 0.073 (the 1.51 denominator is the sum of the length of the rope and the radius of the ball). The weight is then the tension times the cosine of this angle, or w mg (0.270kg)(9.80m s2) T = = = = 2.65 N. cos ? cos(arcsin(.073)) 0.998 The force of the pole on the ball is the tension times sin ? , or (0.073)T = 0.193 N.

5.13: a) In the absence of friction, the force that the rope between the blocks exerts on block B will be the component of the weight along the direction of the incline, T = w sin ? . b) The tension in the upper rope will be the sum of the tension in the lower rope and the component of block A’s weight along the incline, w sin ? + w sin ? = 2w sin ?. c) In each case, the normal force is w cos ?. d) When ? = 0, n = w, when ? = 90??, n = 0.

vertical. At constant speed, the net force is zero, and so F = f and w = L. b) When the plane attains the new constant speed, it is again in equilibrium and so the new values of thethrustanddrag,F?andf?,arerelatedbyF?=f?;ifF?=2F,f?=2f.c)Inorderto increase the magnitude of the drag force by a factor of 2, the speed must increase by a factor of 2 .

5.15: a) The tension is related to the masses and accelerations by T-mg=ma 1 11 2 22 b) For the bricks accelerating upward, let a = -a = a (the counterweight will 12 accelerate down). Then, subtracting the two equations to eliminate the tension gives (m -m )g = (m + m )a, or 2112 m – m ? 28.0 kg – 15.0 kg ? m + m ? 28.0 kg + 15.0kg ? 21 c) The result of part (b) may be substituted into either of the above expressions to find the tension T = 191 N. As an alternative, the expressions may be manipulated to eliminate a algebraically by multiplying the first by m and the second by m and adding 21 (with a = -a ) to give 21 T (m + m ) – 2m m g = 0, or 1 2 12 2m m g 2(15.0 kg) (28.0 kg) (9.80 m s2 ) m + m (15.0 kg + 28.0 kg) 12 In terms of the weights, the tension is 2m 2m m +m m +m 12 1212 If, as in this case, m > m , 2m > m + m and 2m < m + m , so the tension is greater 21212112 than w and less than w ; this must be the case, since the load of bricks rises and the 12 counterweight drops.

g sin ? = v2 2x, or 5.17: a) b) In the absence of friction, the net force on the 4.00-kg block is the tension, and so theaccelerationwillbe(10.0N)(4.00kg)=2.50ms2.c)Thenetupwardforceonthe suspended block is T – mg = ma, or m = T (g + a). The block is accelerating downward, 5.18: The maximum net force on the glider combination is 12,000 N – 2 ?ù 2500 N = 7000 N, max 1400 kg

a) In terms of the runway length L and takoff speed v, a = v < a , so 2

2 L max v2 (40 m s)2 2a 2(5.0 m s2 ) max b) If the gliders are accelerating at a , from max T – F = ma, T = ma + F = (700 kg)(5.0 m s2 ) + 2500 N = 6000 N. Note that this is drag drag exactly half of the maximum tension in the towrope between the plane and the first glider.

5.19: Denote the scale reading as F, and take positive directions to be upward. Then, w ?F ? g ?w ? 22

gt 2 terms of velocity along the table, t = x , x being the length of the table and v the v0 0

velocity component along the table. Then, ? 2L ? ? 2Lv2 ? arcsin? ? = arcsin? 0 ? ? g(x v ) ? gx ? 22 ? 0? ?2(2.50?ù10-2m)(3.80ms)2? ? m s2 (1.75 m)2 ?

box in equilibrium. b) The maximum static friction force is, from Eq. (5.6), ?Á n = ?Á w = (0.40) (40.0 N) = 16.0 N, so the box will not move and the friction force ss balances the applied force of 6.0 N. c) The maximum friction force found in part (b), 16.0 N. d) From Eq. (5.5), ?Á n = (0.20)(40.0 N) = 8.0 N e) The applied force is enough k

to either start the box moving or to keep it moving. The answer to part (d), from Eq. (5.5), is independent of speed (as long as the box is moving), so the friction force is k

5.25: a) At constant speed, the net force is zero, and the magnitude of the applied force must equal the magnitude of the kinetic friction force, r kkk

b) r F – f = ma, so k r F = ma + f = ma = ?Á mg = m(a + ?Á g) kkk

f 5.26: The coefficient of kinetic friction is the ratio k , and the normal force has n

magnitude 85 N + 25 N = 110 N. The friction force, from F – f = ma = w a is Hk g

a ? – 0.9 m s2 ? ? 2 ? = f = F – w = 20 N – 85 N 28 N kH g ? 9.80 m s ? k 110 N

v2 v2 (28.7 m s)2 2a 2?Á g 2(0.80)(9.80m s2) k

b) The stopping distance is inversely proportional to the coefficient of friction and proportional to the square of the speed, so to stop in the same distance the initial speed should not exceed ?Á 0.25 ?Á 0.80 k,dry

5.29: For a given initial speed, the distance traveled is inversely proportional to the coefficient of kinetic friction. From Table 5.1, the ratio of the distances is then 0.44 = 11. 0.04

5.30: (a) If the block descends at constant speed, the tension in the connecting string must be equal to the hanging block’s weight, w . Therefore, the friction force ?Á w on B kA B B kA BkABA

5.31: a) For the blocks to have no acceleration, each is subject to zero net force. Considering the horizontal components, r T = f , F = T + f , or AB

a v2 – v2 v2 – 1 v2 3 v2 ?Á==0=040=0, r g 2Lg 2Lg 8 Lg where L is the distance covered before the wheel’s speed is reduced to half its original speed.Lowpressure,L=18.1m;3 (3.50ms)2 =0.0259.Highpressure, 8 (18.1 m)(9.80 m s 2 )

L = 3 (3.50 m s)2 = 8 (92.9m)(9.80m s2) 5.33: Without the dolly: n = mg and F – ?Á n = 0 ( a = 0 since speed is constant). kx F 160 N m = = = 34.74 kg ?Á g (0.47)(9.80m s2) k

With the dolly: the total mass is 34.7 kg + 5.3 kg = 40.04 kg and friction now is rolling rr F – ?Á mg = ma r

F – ?Á mg a = r = 3.82 m s2 m 5.34: Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and F in the horizontal direction must be zero. Therefore f = ?Á n = net r r = F = 200 N before the weight and pressure changes are made. After the changes, horiz (0.81) (1.42n) = F , because the speed is still constant and F = 0 . We can simply horiz net divide the two equations: (0.81?Á )(1.42n) r = Fhoriz r

There are two equations and two unknowns, a and T: -?Ámg+T=ma kA A mg-T=ma BB B kA B A (b) Solving either equation for the tension gives T = 11.7 N.

5.36: a) The normal force will be w cos ? and the component of the gravitational force alongtherampiswsin?.Theboxbeginstoslipwhenwsin?>?Áwcos?,or s

tan ? > ?Á = 0.35, so slipping occurs at ? = arctan(0.35) = 19.3?? , or 19?? to two figures. s

b) When moving, the friction force along the ramp is ?Á w cos ? , the component of the k

gravitational force along the ramp is wsin ? , so the acceleration is kk (c)2ax=v2,sov=(2ax)12,orv=[(2)(0.92ms2)(5m)]12=3m.

5.37: a) The magnitude of the normal force is mg + r F r of F , r F cos?

solving for r F must balance the frictional force, so r F cos? = ?Á (mg + k r F

gives r F ?Á mg =k cos? – ?Á sin? k sin sin ?. The horizontal component

b) If the crate remains at rest, the above expression, with ?Á instead of ?Á , gives the sk force that must be applied in order to start the crate moving. If cot ? < ?Á , the needed s

n=w-Fsin?=mg-Fsin?.Thefrictionforceisf=?Án=?Á(mg-Fsin?).Thenet kkk horizontalforceisFcos?-f=Fcos?-?Á(mg-Fsin?),andsoatconstantspeed, kk ?Á mg F= k cos?+?Á sin? k

b) Using the given values, (0.35)(90kg)(9.80m s2) F = = 293 N, (cos 25?? + (0.35) sin 25??) or 290 N to two figures.

5.39: a) b) The blocks move with constant speed, so there is no net force on block A; the tension in the rope connecting A and B must be equal to the frictional force on block A, ?Á = (0.35) (25.0 N) = 9 N. c) The weight of block C will be the tension in the rope k

connecting B and C; this is found by considering the forces on block B. The components of force along the ramp are the tension in the first rope (9 N, from part (a)), the component of the weight along the ramp, the friction on block B and the tension in the second rope. Thus, the weight of block C is w=9N+w(sin36.9??+?Ácos36.9??) CBk = 9 N + (25.0 N)(sin 36.9?? + (0.35)cos 36.9??) = 31.0 N, or 31 N to two figures. The intermediate calculation of the first tension may be avoided to obtain the answer in terms of the common weight w of blocks A and B, w =w(?Á +(sin?+?Á cos?)), Ckk giving the same result.

? k ? -( ) -( ) a = v ? ?e k m t = ge k m t , t?m? t

Integrating Eq. (5.10) with respect to time with y = 0 gives 0 t

= ? – -(k m)t y v [1 e ] dt t 0 ? ? m ? -(k m ? ? m ? = v ?t + ? ?e )t ? – v ? ? t? ?k? ? t? ? k

(1 m)t )? ?m- (k t? k ? 5.41: a) Solving for D in terms of v , t

mg (80 kg) (9.80 m s2 ) v2 (42 m s)2 t mg (45 kg)(9.80 m s2 ) b) v = = = 42 m s. t D (0.25 kg m)

5.42: At half the terminal speed, the magnitude of the frictional force is one-fourth the weight. a) If the ball is moving up, the frictional force is down, so the magnitude of the net force is (5/4)w and the acceleration is (5/4)g, down. b) While moving down, the frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration is (3/4)g, down.

5.43: Setting F equal to the maximum tension in Eq. (5.17) and solving for the speed v net gives F R (600 N)(0.90 m) v = net = = 26.0 m s , m (0.80 kg) or 26 m/s to two figures.

5.44: This is the same situation as Example 5.23. Solving for ?Á yields s

that the net vertical force is zero, so F cos ? = (3.8)w cos ? = w, , and (b) The angle does not depend on speed.

5.46: a) The analysis of Example 5.22 may be used to obtain tan? = (v2 gR), but the subsequent algebra expressing R in terms of L is not valid. Denoting the length of the horizontal arm as r and the length of the cable as l, R = r + l sin ?. The relation v = 2?R is T

? = 4? R = 4? (r +l sin ? ) 22 still valid, so tan . Solving for the period T, 22 gT gT

4?2 (r + l sin ?) 4?2 (3.00 m + (5.00 m)sin 30??) T = = = 6.19 s. g tan ? (9.80 m s2 ) tan 30?? Note that in the analysis of Example 5.22, ? is the angle that the support (string or cable) makes with the vertical (see Figure 5.30(b)). b) To the extent that the cable can be considered massless, the angle will be independent of the rider’s weight. The tension in the cable will depend on the rider’s mass.

5.47: This is the same situation as Example 5.22, with the lift force replacing the tension in the string. As in that example, the angle ? is related to the speed and the turning radius by tan ? = v . Solving for ? , 2

gR ? v2 ? ? (240 km h ?ù ((1m s) (3.6 km h)))2 ? ? gR ? ? 9.80 m s2 1200 m) ?

5.48: a) This situation is equivalent to that of Example 5.23 and Problem 5.44, so ?Á = v . Expressing v in terms of the period T, v = 2?R , so ?Á = 4? R . A platform speed 22

rad R 400 m T = 2? = 2? = 40.1s, g 9.80 m s2 b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square root of the ratio of the accelerations, T ? = (1.5 rev min) ?ù 3.70 9.8 = 0.92 rev min. .

5.50: a) 2? R T = 2?(50.0 m) (60.0 s) = 5.24 m s. b) The magnitude of the radial force is mv2 R = m4? 2 R T 2 = w(4? 2 R gT 2 ) = 49 N (to the nearest Newton), so the apparent weight at the top is 882 N – 49 N = 833 N, and at the bottom is 882 N + 49 N = 931 N . c) For apparent weightlessness, the radial acceleration at the top is equal to g in magnitude. Using this in Eq. (5.16) and solving for T gives R 50.0 m g 9.80 m s2 d) At the bottom, the apparent weight is twice the weight, or 1760 N.

5.51: a) If the pilot feels weightless, he is in free fall, and a = g = v2 R , so v=Rg=(150m)(9.80ms2)=38.3ms,or138kmh.b)Theapparentweightis the sum of the net inward (upward) force and the pilot’s weight, or ? a? w + ma = w ?1 + ? ? g? ( )? (280 km h) ? 2

= 700 N ?1 + ? ? (3.6(kmh)(ms))2(9.80ms2)(150m)?? = 3581 N, or 3580 N to three places.

5.52: a) Solving Eq. (5.14) for R, b) The apparent weight will be five times the actual weight, 5mg=5(50.0kg)(9.80ms2)=2450N to three figures.

5.54: a) The inward (upward, radial) acceleration will be v2 = (4.2m s)2 = 4.64 2 m s . At the R (3.80 m) b) The forces on the ball are tension and gravity, so T – mg = ma, ? a ? ? 4.64 m s2 ? ? g ? ? 9.80 m s2 ? 5.55: a)

The tension in the lower chain balances the weight and so is equal to w. The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w, and so the tension in the rope is w 2 . Then, the downward force on the upper pulley due to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w.

5.57: In the absence of friction, the only forces along the ramp are the component of the r weight along the ramp, w sin? , and the component of F along the ramp, r Fcos?=Fcos?.Theseforcesmustsumtozero,soF=wtan?.

Considering horizontal and vertical components, the normal force must have horizontal component equal to nsin? , which must be equal to F; the vertical component must balance the weight, n cos? = w . Eliminating n gives the same result.

5.58: The hooks exert forces on the ends of the rope. At each hook, the force that the hook exerts and the force due to the tension in the rope are an action-reaction pair. The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts an upward vertical force of w 2 on the rope. Therefore, the downward force that end end b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component of the force due to the tension at the end; T cos ? = T , so end middle middle (c) Mathematically speaking, ? ? 0 because this would cause a division by zero in the equation for T or T . Physically speaking, we would need an infinite tension to end middle keep a non-massless rope perfectly straight.

pointareTupand(M+m(L-x))gdownwards.Newton’sSecondLawbecomes L

T-(M+m(L-x))g=(M+m(L-x))a.Sincea=F-(M+m)g,T=(M+m(L-x))(F)..At L L M+m L M+m M +m

5.60: a) The tension in the cord must be m g in order that the hanging block move at 2

constant speed. This tension must overcome friction and the component of the gravitational force along the incline, so m g = (m g sin ? + ?Á m g cos ?) and 2 1 k1 21 k b) In this case, the friction force acts in the same direction as the tension on the block of mass m , so m g = (m g sin ? – ?Á m g cos ?) , or m = m (sin ? – ?Á cos? ) . 1 2 1 k1 2 1 k c) Similar to the analysis of parts (a) and (b), the largest m could be is 2

1s21s 5.61: For an angle of 45.0?? , the tensions in the horizontal and vertical wires will be the same. a) The tension in the vertical wire will be equal to the weight w = 12.0 N ; this must be the tension in the horizontal wire, and hence the friction force on block A is also 12.0 N . b) The maximum frictional force is ?Á w = (0.25)(60.0 N) = 15 N ; this will be the sA tension in both the horizontal and vertical parts of the wire, so the maximum weight is 15 N.

5.62: a) The most direct way to do part (a) is to consider the blocks as a unit, with total weight 4.80 N. Then the normal force between block B and the lower surface is 4.80 N, and the friction force that must be overcome by the force F is ?Á n = (0.30)(4.80 N) = 1.440 N, or 1.44 N, to three figures. b) The normal force k

r F by F.) a) The force normal to the surface is n = F cos ? ; the vertical component of the applied force must be equal to the weight of the brush plus the friction force, so that F sin? = w + ?Á F cos ? , and k

w 12.00 N F = = = 16.9 N, sin?-?Á cos? sin53.1??-(0.51)cos53.1?? k

5.64: a) ?F=ma=m(62.5g)=62.5mg =(62.5)(210?ù10-6g)(980cms2) =13dynes=1.3?ù10-4 N b) F =ma =m(140g)=140mg max max =29dynes=2.9?ù10-4 N c) v = v – v = v – 0 = v = area under a-t graph. Approximate area as shown: 0

? F = ma yy T – mg = ma T – mg a = = 13.07 m s2 m 0y y y v = v + a t gives t = 25.3 s y 0y y Consider forces on the rocket; rocket has the same a . Let F be the thrust of the y

F – mg = ma F=m(g+a)=(25,000kg)(9.80ms2+13.07ms2)=5.72?ù105N 0 0y 2 y 0

? F = ma yy n – mg = ma n = 1.6mg so a = 0.60 g = 5.88 m s2 0 y 0y y

v 2 = v 2 + 2a ( y – y ) gives v = 5.0 m s y 0y y 0 y

5.68: (a) Choosing upslope as the positive direction: F = -mg sin 37?? – f = -mg sin37?? – ?Á mg cos37?? = ma net k k and a=-(9.8ms2)(0.602+(0.30)(0.799))=-8.25ms2 Since we know the length of the slope, we can use v2 = v2 + 2a(x – x ) with x = 0 and 000 v2=-2ax=-2(-8.25ms2)(8.0m)=132m2s2 0

? F = ma gives n = mg cos? yy ? F = ma xx mg sin ? – f = ma k

accelerating, the tension that block A exerts on the rope is different from the tension that block B exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn’t accelerate.) Also, treat the rope as if it is just another object. Taking the ÔÇ£clockwiseÔÇØ direction to be positive, the Second Law equations for the three different parts of the system are: Block A (The only horizontal forces on A are tension to the right, and friction to the left): kA A A Block B (The only vertical forces on B are gravity down, and tension up): BBB Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the portion of the rope that hangs vertically): RL B A R To solve for a and eliminate the tensions, add the left hand sides and right hand sides of the three equations: – ?Á m g + m g + m ( )g = (m + m + m )a, or a = g + -?Á . mmdm d B RL k A k A B R L A B R (m +m +m ) ABR m +m (d ) (a) When ?Á = 0, a = g B R L . As the system moves, d will increase, approaching k (m +m +m ) ABR

L as a limit, and thus the acceleration will approach a maximum value of mm (mA+mB+mR) B RL sA Note that we must use static friction to find d for when the block will begin to move. m s A B .160 kg R

(c) When m = .04 kg, d = 1.0 m (.25(2 kg) – .4 kg) = 2.50 m. This is not a physically R .04 kg possible situation since d > L. The blocks won’t move, no matter what portion of the rope hangs over the edge.

ramp is sss In order to move the box, this net force must be greater than zero. Solving for F, sin ? + ?Á cos ? cos ? – ?Á sin ? s

Since F is the magnitude of a force, F must be positive, and so the denominator of this expressionmustbepositive,orcos?>?Ásin?,and?Á

5.75: a) The product ?Á g = 2.94 m s2 is greater than the magnitude of the acceleration of s

the truck, so static friction can supply sufficient force to keep the case stationary relative to the truck; the crate accelerates north at 2.20 m s2 , due to the friction force of ma = 66.0 N. b) In this situation, the static friction force is insufficient to maintain the case at rest relative to the truck, and so the friction force is the kinetic friction force, kk

5.76: To answer the question, v must be found and compared with 20 m s (72 km hr). 0

The kinematics relationship 2ax = -v2 is useful, but we also need a. The acceleration 0

must be large enough to cause the box to begin sliding, and so we must use the force of static friction in Newton’s Second Law: – ?Á mg ? ma, or a = -?Á g. Then, ss

T cos ? = ?Á (w – T sin ?), max k max and solving for the weight w gives ? cos ? ? max ? ?Ák ? This will be a maximum when the quantity in parentheses is a maximum. Differentiating with respect to ? , d ? cos ? ? sin ? ? + sin ? ? = – + cos ? = 0, d? ? ?Á ? ?Á kk or tan? = ?Á , where ? is the value of ? that maximizes the weight. Substituting for ?Á kk in terms of ? , ? cos ? ? w = T ? + sin ? ? max ? sin ? cos ? ? ? cos2 ? + sin2 ? ? = T ? ? max ? sin ? ? T sin ? b) In the absence of friction, any non-zero horizontal component of force will be enough to accelerate the crate, but slowly.

5.78: a) Taking components along the direction of the plane’s descent, f=wsin?andL=wcos?.b)Dividingoneoftheserelationsbytheothercancelsthe weight, so tan ? = f L. c) The distance will be the initial altitude divided by the tangent 12,900 N This makes the horizontal distance (2500 m) tan(5.78??) = 24.7 km. d) If the drag is reduced, the angle ? is reduced, and the plane goes further.

5.79: If the plane is flying at a constant speed of 36.1 m s, then ? F = 0, or T – wsin? – f = 0. The rate of climb and the speed give the angle ?, ? = arcsin(5 36.1) = 7.96??. Then, T = w sin ? + f . T = (12,900 N) sin 7.96?? + 1300 N = 3087 N. Note that in level flight (? = 0), the thrust only needs to overcome the drag force to maintain the constant speed of 36.1m s.

need to cause an acceleration of 2.20 m s2; however, the maximum acceleration possible duetostaticfrictionis(0.19)(9.80ms2)=1.86ms2,andsotheblockwillmoverelative k

The difference between the distance the truck moves and the distance the box moves (i.e., the distance the box moves relative to the truck) will be 1.80 m after a time

2 x 2(1.80 m) a -a 2- 2 truck box (2.20 m s 1.47 m s )

In this time, the truck moves 1 a t 2 = 1 (2.20 m s 2 ) (2.221 s) 2 = 5.43 m. Note that an 2 truck 2 extra figure was kept in the intermediate calculation to avoid roundoff error.

5.81: The friction force on block A is ?Á w = (0.30)(1.40 N) = 0.420 N, as in Problem 5- kA 68. This is the magnitude of the friction force that block A exerts on block B, as well as the tension in the string. The force F must then have magnitude F = ?Á (w + w ) + ?Á w + T = ?Á (w + 3w ) kBAkAkBA = (0.30)(4.20 N + 3(1.40 N)) = 2.52 N.

Note that the normal force exerted on block B by the table is the sum of the weights of the blocks.

– 3.7 m s2 for the first 20 s. Thus at the end of that time her vertical velocity will be v = at = (-3.7 m s2 )(20 s) = -74 m s. She will have fallen a distance y

? – 74 m s ? d = v t = ? ?(20 s) = -740 m av ? 2 ?

and will thus be 1200 – 740 = 460 m above the surface. Her vertical velocity must reach zero as she touches the ground; therefore, taking the ignition point of the PAPS as y = 0, 0

v2=v2+2a(y-y) y0 0 v2 – v2 – – 2 y 0 0 ( 74 m s) a = = = 5.95 m s2 or 6.0 m s2 2( y – y ) – 460 0

which is the vertical acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by v – v 0 – (-74 m s) t = 0 = = 12.4 s a 5.95 m s 2

Using Newton’s Second Law for the vertical direction F + mg = ma PAPSv F =ma-mg=m(a+g)=(150kg)(5.95-(-3.7))ms2 PAPSv = 1447.5 N or 1400 N which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in 12.4 seconds; the acceleration needed to do this is v – v 0 – 33 m s 2 a = 0 = = 2.66 m s 2 t 12.4 s andtheforceneededisF =ma=(150kg)(2.66ms2)=399Nor400N,sincethere PAPSh are no other horizontal forces.

A attached to block C be T . The equations of motion are then C

T-mg=ma AAA T-?Ámg-T=ma C kB A B CCC a) Adding these three equations to eliminate the tensions gives a(m + m + m ) = g(m – m – ?Á m ), A B C C A kB solving for m gives C

m (a + g) + m (a + ?Á g) m=A B k, C g-a C

AA CC 5.84: Considering positive accelerations to be to the right (up and to the right for the left- hand block, down and to the right for the right-hand block), the forces along the inclines and the accelerations are related by T-(100kg)gsin30??=(100kg)a,(50kg)gsin53??-T=(50kg)a,whereTisthetension in the cord and a the mutual magnitude of acceleration. Adding these relations, (50kgsin53??-100kgsin30??)g=(50kg+100kg)a,ora=-0.067g.a)Sinceacomes out negative, the blocks will slide to the left; the 100-kg block will slide down. Of course, if coordinates had been chosen so that positive accelerations were to the left, a would be c) Substituting the value of a (including the proper sign, depending on choice of coordinates) into either of the above relations involving T yields 424 N.

1 of mass m will descend with acceleration a 2. If the tension in the rope is T, the 2

equations of motion are then T=ma 1 22 Multiplying the first of these by 2 and adding to eliminate T, and then solving for a gives m g 2m 2m + m 2 4m + m 12 12

The acceleration of the block of mass m is half of this, or g m (4m + m ). 2 212

5.86: Denote the common magnitude of the maximum acceleration as a. For block A to remain at rest with respect to block B, a < ?Á g. The tension in the cord is then s

T = (m + m )a + ?Á g(m + m ) = (m + m )(a + ?Á g). This tension is related to the mass ABkABABk m by T = m (g – a). Solving for a yields CC

m – ?Á (m + m ) m +m +m s ABC Solving the inequality for m yields C

(m + m )(?Á + ?Á ) C -?Á 1 s 5.87: See Exercise 5.15 (Atwood’s machine). The 2.00-kg block will accelerate upward at g – = 3g 7 , and the 5.00-kg block will accelerate downward at 3g 7. Let the 5.00 kg 2.00 kg 5.00 kg + 2.00 kg initial height above the ground be h ; when the large block hits the ground, the small 0

block will be at a height 2h , and moving upward with a speed given by 0

v2=2ah=6gh7.Thesmallblockwillcontinuetoriseadistancev22g=3h7,and 000 00 so the maximum height reached will be 2h + 3h 7 = 17h 7 = 1.46 m , which is 0.860 m 000 above its initial height.

n = m(a + g). The friction force that needs to be balanced is kk

5.89: The upward friction force must be f = ?Á n = m g, and the normal force, which is ssA the only horizontal force on block A, must be n = m a, and so a = g ?Á . An observer on As the cart would ÔÇ£feelÔÇØ a backwards force, and would say that a similar force acts on the block, thereby creating the need for a normal force.

5.90: Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be pulled down the plane; i.e., the larger block will not move faster than the smaller block, and the blocks will have the same acceleration. For the smaller block, (4.00kg)g(sin30??-(0.25)cos30??)-T=(4.00kg)a,or11.11N-T=(4.00kg)a,and similarly for the larger, 15.44 N + T = (8.00 kg)a, a) Adding these two relations, 26.55N=(12.00kg)a,a=2.21ms2(notethatanextrafigurewaskeptinthe intermediate calculation to avoid roundoff error). b) Substitution into either of the above relations gives T = 2.27 N. Equivalently, dividing the second relation by 2 and subtracting from the first gives 3 T = 11.11 N – 15.44 N , giving the same result. c) The 22 string will be slack. The 4.00-kg block will have a = 2.78 m s2 and the 8.00-kg block will have a = 1.93 m s2 , until the 4.00-kg block overtakes the 8.00-kg block and collides with it.

5.91: a) Let n be the normal force between the plank and the block and n be the BA normal force between the block and the incline. Then, n = wcos? and B

n = n + 3w cos ? = 4w cos ?. The net frictional force on the block is AB ?Á (n + n ) = ?Á 5w cos? . To move at constant speed, this must balance the component kABk oftheblock’sweightalongtheincline,so3wsin?=?Á5wcos?,and k

The equation of motion for the man is T + n – Mg = Ma, where T is the tension in the rope, and for the platform, T – n – mg = ma . Adding to eliminate n, and rearranging, T = 1 (M + m)(a + g). This result could be found directly by considering the man- 2

platform combination as a unit, with mass m + M , being pulled upward with a force 2T due to the two ropes on the combination. The tension T in the rope is the same as the force that the man applies to the rope. Numerically, 1 2

2 (b) The end of the rope moves downward 2 m when the platform moves up 1 m, so a = -2a . Relative to the man, the acceleration of the rope is 3a = 5.40 m s2, rope platform downward.

5.93: a) The only horizontal force on the two-block combination is the horizontal r component of F , F cos ?. The blocks will accelerate with a = F cos? (m + m ). b) The 12 normal force between the blocks is m g + Fsin? , for the blocks to move together, the 1

product of this force and ?Á must be greater than the horizontal force that the lower block s

exerts on the upper block. That horizontal force is one of an action-reaction pair; the reaction to this force accelerates the lower block. Thus, for the blocks to stay together, ma??Á(mg+Fsin?).Usingtheresultofpart(a), 2s1 Fcos? m +m 2 s1 12 Solving the inequality for F gives the desired result.

tan ? = v2 , where v is the ideal speed, 20 m s in this case. For speeds larger than v , a 0 gR 0 0 frictional force is needed to keep the car from skidding. In this case, the inward force will consist of a part due to the normal force n and the friction force f ; n sin ? + f cos ? = ma . The normal and friction forces both have vertical rad components; since there is no vertical acceleration, n cos ? – f sin ? = mg. Using f=?Ánanda =v2=(1.5v0)2=2.25gtan?,thesetworelationsbecome s rad R R nsin?+?Áncos?=2.25mgtan?, s

s Dividing to cancel n gives sin ? + ?Á cos ? cos ? – ?Á sin ? s

Solving for ?Á and simplifying yields s ?Á 1.25 sin? cos ? 1+1.25sin2 ? s 2 (20 m s) (9.80 m s )(120 m) s 2

5.95: a) The same analysis as in Problem 5.90 applies, but with the speed v an unknown. The equations of motion become nsin?+?Áncos?=mv2R, s

s Dividing to cancel n gives sin ? + ?Á cos ? v2 cos ? – ?Á sin ? Rg s

Solving for v and substituting numerical values gives v = 20.9 m s (note that the value b) The same analysis applies, but the friction force must be directed up the bank; this has the same algebraic effect as replacing f with – f , or replacing ?Á with – ?Á ss (although coefficients of friction may certainly never be negative). The result is sin ? – ?Á cos ? v 2 = ( gR) s , cos ? + ?Á sin ? s

the road is provided by friction; thus ?Á mg = mv 2 sr v 2 (35.7 m s) 2 r = = = 171 m or 170 m ?Á g (0.76)(9.8 m s 2 ) s

(b) If ?Á = 0.20 : s v2=r?Ág=(171m)(0.20)(9.8ms2)=335.2m2s2 s v = 18.3 m s or about 41 mi h (c) If ?Á = 0.37 : s

v2=(171m)(0.37)(9.8ms2)=620m2s2 v = 24.9 m s or about 56 mi h The speed limit is evidently designed for these conditions.

5.97: a) The static friction force between the tires and the road must provide the centripetal acceleration for motion in the circle.

?Á mg = m v2 s r vv m, g, and r are constant so 1 = 2 , where 1 refers to dry road and 2 to wet ?Á?Á s1 s 2 ?Á=1?Á,sov=(27ms)2=19ms s 2 2 s1 2 b) Calculate the time it takes you to reach the curve 0x x 0 ?v +v ? x – x = ? 0x x ? t gives t = 34.7 s 0?2? During this time the other car will travel x – x = v t = (36 m s) (34.7s) = 1250 m. The 0 0x other car will be 50 m behind you as you enter the curve, and will be traveling at nearly twice your speed, so it is likely it will skid into you.

weight, and the tension in the string is the same at the point where the bananas are suspended and where the monkey is pulling; in all cases, the monkey and bananas will have the same net force and hence the same acceleration, direction and magnitude. a) The bananas move up. b) The monkey and bananas always move at the same velocity, so the distance between them stays the same. c) Both the monkey and bananas are in free fall, and as they have the same initial velocity, the distance bewteen them doesn’t change. d) The bananas will slow down at the same rate as the monkey; if the monkey comes to a stop, so will the bananas.

5.100: The separated equation of motion has a lower limit of 3v instead of 0; t

specifically, v v 2v 2v 2 m 3vt t t t

0 b) (18.0 N – (2.20 N ? s m) (3.00 m s)) (3.00 kg) = 3.80 m s2. c) The net force must be 1.80 N, so kv = 16.2 N and v = (16.2 N) (2.20 N ? s m) = 7.36 m s. d) When the net force is equal to zero, and hence the acceleration is zero, kv = 18.0 N and t

v = (18.0 N) (2.20 N ? s m) = 8.18 m s. e) From Eq. (5.12), t

? 3.00 kg – ? ? y = (8.18 m s)?(2.00 s) – (1 – e )? ((2.20 N s m) (3.00 kg))(2.00 s)

? 2.20 N ? s m ? From Eq. (5.10), = – -((2.2N?sm)(3.00kg))(2.00s) v (8.18 m s)[1 e ] From Eq. (5.11), but with a instead of g, 0

= 2 -((2.20N?sm)(3.00kg))(2.00s)= 2 a (6.00 m s )e 1.38 m s .

F F – kv1 2 dv a = net = R = = m m m dt dv k = – dt v1 2 m vt

? dv k ? = dt v1 2 m v0 0

2v1 2 which gives v=v – 0 For the rock’s position: dx =v – 0 v1 2 0

dt m 4m v1 2ktdt k 2t 2dt dx = v dt – 0 + 0 m 4m2 kt v =- v0 m

mg mg t v 0.36 s t With buoyancy included there is the additional upward buoyancy force B, so B – kv = mg t

? 0.24 m s ? B = mg – kv = mg?1 – ? = mg 3 t ? 0.36 m s ?

5.104: Recognizing the geometry of a 3-4-5 right triangle simplifies the calculation. For a) Balancing the vertical force, T 4 – T 4 = w, so U5 L5 55 LU 44 b) The net inward force is F = 3 T + 3 T = 66.6 N. Solving F = ma = m 4? R for 2

5 U 5 L rad T 2 the period T, mR (4.00 kg) (0.75 m) T = 2? = 2? = 1.334 s, F (66.6 N) or 0.02223 min, so the system makes 45.0 rev/min. c) When the lower string becomes slack, the system is the same as the conical pendulum considered in Example 5.22. With cos ? = 0.800, the period is T = 2? (1.25 m) (0.800) (9.80 m s2 ) = 2.007 s, which is the same as 29.9 rev min. d) The system will still be the same as a conical pendulum, but the block will drop to a smaller angle.

dv mg m = mg – kv , where = v y yt dt k

v? dvy = – k ?t dt y v -v m v0 y t 0 This is the same expression used in the derivation of Eq. (5.10), except the lower limit 0

Evaluating the integrals and rearranging gives = -kt m + – -kt m v v e v (1 e ) 0t Note that at t = 0 this expression says v = v and at t ? ? it says v ? v . y0 yt b) The downward gravity force is larger than the upward fluid resistance force so the acceleration is downward, until the fluid resistance force equals gravity when the terminal speed is reached. The object speeds up until v = v . Take + y to be downward. yt

c) The upward resistance force is larger than the downward gravity force so the acceleration is upward and the object slows down, until the fluid resistance force equals gravity when the terminal speed is reached. Take + y to be downward.

v2=v2+2a(y-y) y 0y 0

v2 – v2 0 – (6.0 m s)2 y – y = y 0 y = = 1.84 m or 1.8 m 0 2a 2(-9.8 m s2 ) v – v 0 – 6.0 m s t = 0 = = 0.61s a – 9.8 m s2 (b) Starting from Newton’s Second Law for this situation dv m = mg – kv dt we rearrange and integrate, taking downward as positive as in the text and noting that the velocity at the top of the rock’s ÔÇ£flightÔÇØ is zero: 0

? dv k =- t v-v m vt – v – 2.0 m s ln(v – v ) 0 = ln t = ln = ln(0.25) = -1.386 t v v – v – 6.0 m s – 2.0 m s t

FromEq.5.9,mk=vg=(2.0ms2)(9.8ms2)=0.204s,and t t = – m (-1.386) = (0.204 s) (1.386) = 0.283 s to the top. Equation 5.10 in the text gives us k

dx = v (1 – e-(k m)t = – -(k m)t ) v ve t tt dt xtt

? = ? – ? -(k m)t dx v dt v e dt tt 000

D force ?Á mg. Take the velocity to be in the + x -direction. The forces are opposite in r

direction to the velocity. ? F = ma gives xx – Dv2 – ?Á mg = ma r

We can write this equation twice, once with v = 32 m s and a = – 0.42 m s2 and once with v = 24 m s and a = -0.30 m/s2. Solving these two simultaneous equations in the rr b) n = mg cos ? and the component of gravity parallel to the incline is mg sin ?, r

c) For angle ? , mg sin ? – ?Á mg cos ? – Dv2 = 0 r mg(sin ? – ?Á cos ?) and v = r D The terminal speed for a falling object is derived from Dv2 – mg = 0, so t

t v v = sin ? – ?Á cos ? tr And since ?Á = 0.015, v v = sin ? – (0.015) cos ? rt

5.108: (a) One way of looking at this is that the apparent weight, which is the same as the upward force on the person, is the actual weight of the person minus the centripetal force needed to keep him moving in its circular path: mv2 ? (12 m s)2 ? w = mg – = (70 kg) ?(9.8 m s ) – ? 2 app R ? 40 m ? = 434 N (b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when the road no longer has to exert any upward force on it: mv2 mg – = 0 R v2=Rg=(40m)(9.8ms2)=392m2s2 v = 19.8 m s or 20 m s The answer doesn’t depend on the cart’s mass, because the centripetal force needed to hold it on the road is proportional to its mass and so is its weight, which provides the centripetal force in this situation.

proportional to the radius, and for twins of the same mass, the needed force is proportional to the radius; Jackie is twice as far away from the center, and so must hold Jackie

30 kg 5.110: The passenger’s velocity is v = 2? R t = 8.80 m s. The vertical component of the seat’s force must balance the passenger’s weight and the horizontal component must provide the centripetal force. Therefore: F sin ? = mg = 833 N seat mv2 F cos ? = = 188 N seat R Therefore tan ? = 833 N 188 N = 4.43;? = 77.3?? above the horizontal. The magnitude of the net force exerted by the seat (note that this is not the net force on the passenger) is ? mv2 ?2 F = (mg)2 + ? ? = (833 N)2 + (188 N)2 seat ?R? = 854 N (b) The magnitude of the force is the same, but the horizontal component is reversed.

5.111: a) b) The upward friction force must be equal to the weight, so ?Án=?Ám(4?2RT2)?mgand ss

22 s 4?2R 4?2(2.5m) c) No; both the weight and the required normal force are proportional to the rider’s mass.

Thus, the (downward) acceleration at the top of the sphere must exceed mg, so 2

R b) The (upward) acceleration will then be 4g, so the upward normal force must be 5mg=5(110kg)(9.80ms2)=5390N.

5.113: a) What really happens (according to a nosy observer on the ground) is that you slide closer to the passenger by turning to the right. b) The analysis is the same as that of Example 5.23. In this case, the friction force should be insufficient to provide the inward radial acceleration, and so ?Á mg < mv2 R, or s

v2 (20 m s)2 R < = = 120 m ?Á g (0.35) (9.80 m s2 ) s

5.114: The tension F in the string must be the same as the weight of the hanging block, and must also provide the resultant force necessary to keep the block on the table in 2

r 5.115: a) The analysis is the same as that for the conical pendulum of Example 5.22, and so ? gT 2 ? ? (9.80 m s2 )(1 4.00 s)2 ? ?4?2L? ? 4?2(0.100m) ? b) For the bead to be at the same elevation as the center of the hoop, ? = 90?? and cos ? = 0, which would mean T = 0, the speed of the bead would be infinite, and this is not possible. c) The expression for cos ? gives cos ? = 2.48, which is not possible. In deriving the expression for cos ?, a factor of sin ? was canceled, precluding the possibility that ? = 0. For this situation, ? = 0 is the only physical possibility.

xy F = ma = (2.20 kg) (-0.72 N s)t = -(1.58 N/s)t xx yy b) c) At t = 3.00 s, F = -4.75 N and F = -4.40 N, so xy

F = (-4.75 N)2 + (-4.40 N)2 = 6.48 N, -4.75 5.117: 5.118: See Example 5.25. ( ) 2 (12.0 m s ) 2 b)FA=m(g-)=(1.60kg)(9.80ms- )=-30.4N.,wheretheminussign R 5.00 m v2 2 (12.0ms)2 B R 5.00 m indicates that the track pushes down on the car. The magnitude of this force is 30.4 N.

related to the period by v = 2? R T = 2?h tan ? T , or T = 2?h tan ? v. The maximum and minimum speeds are the same as those found in Problem 5.95, cos ? + ?Á sin ? v = gh tan ? s sin ? – ?Á cos ? max s

cos ? – ?Á sin ? min sin ? + ?Á cos ? s The minimum and maximum values of the period T are then h tan ? sin ? – ?Á cos ? T = 2? s g cos ? + ?Á sin ? min s

fairly straightforward in terms of application of Newton’s laws, but involves a good deal of algebra. For both parts, take the x-direction to be horizontal and positive to the right, and the y-direction to be vertical and positive upward. The normal force between the block and the wedge is n; the normal force between the wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The horizontal acceleration of the wedge is A, and the components of acceleration of the block are a x

and a . The equations of motion are then y MA = -n sin ? ma = n sin ? x

y Note that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are three equations in four unknowns, A, a , a and n. Solution xy is possible with the imposition of the relation between A, a and a . xy An observer on the wedge is not in an inertial frame, and should not apply Newton’s laws, but the kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical acceleration of the block is a , but the y

horizontal acceleration of the block is a – A. To this observer, the block descends at an x

angle ? , so the relation needed is a a -A x At this point, algebra is unavoidable. Symbolic-manipulation programs may save some solution time. A possible approach is to eliminate a by noting that a = – M A (a result x xm that anticipates conservation of momentum), using this in the kinematic constraint to eliminate a and then eliminating n. The results are: y

– gm A= (M + m) tan ? + (M tan ?) gM a= (M + m) tan ? + (M tan ?) x

– g(M + m) tan ? a= y (M + m) tan ? + (M tan ?) (b) When M >> m, A ? 0, as expected (the large block won’t move). Also, a ? ? ? = g ? = g sin ? cos ?, which is the acceleration of the block g tan x tan +(1 tan ) tan ? +1 2

( g sin ? in this case), with the factor of cos ? giving the horizontal component. Similarly, y

common acceleration is a = g tan ?, so the applied force must be (M + m)a = (M + m)g tan ?.

5.122: The normal force that the ramp exerts on the box will be n = w cos ? – T sin ?. The rope provides a force of T cos ? up the ramp, and the component of the weight down the ramp is w sin ?. Thus, the net force up the ramp is F=Tcos?-wsin?-?Á(wcos?-Tsin?) k

kk The acceleration will be the greatest when the first term in parantheses is greatest; as in k

kk b) c) The expression for F is a minimum when the denominator is a maximum; the calculus is identical to that of Problem 5.77 (maximizing w for a given F gives the same result as minimizing F for a given w), and so F is minimized at tan ? = ?Á . For k

released from rest, the acceleration and velocity will be positive, and the speed of the baseball is the same as its positive component of velocity. Then the resisting force, a)

b) Newton’s Second Law is then ma = mg – Dv2. Initially, when v = 0, the acceleration is g, and the speed increases. As the speed increases, the resistive force increases and hence the acceleration decreases. This continues as the speed approaches the terminal speed. c) At terminal velocity, a = 0, so v = , in agreement with mg tD Eq.(5.13). d)Theequationofmotionmayberewrittenasdv=g(v2-v2).Thisisa dt v t2 t separable equation and may be expressed as ? dv g ? = dt, or v2 – v2 v2 tt

1 ? v ? gt arctanh? ? = , vt ? vt ? v 2t tt Note: If inverse hyperbolic functions are unknown or undesirable, the integral can be done by partial fractions, in that 1 1?1 1? = ? + ?, v2-v2 2v?v-v v+v? t tt t and the resulting logarithms in the integrals can be solved for v(t) in terms of exponentials.

straightforward, but the kinematic relations between the accelerations, and the resultant algebra, are not immediately obvious. If the acceleration of pulley B is a , then B

a = -a , and a is the average of the accelerations of masses 1 and 2, or B3B a + a = 2a = -2a . There can be no net force on the massless pulley B, so T = 2T . 12B3 CA The five equations to be solved are then mg-T =ma 1 A 11 mg-T=ma 2 A 22 mg-T =ma 3 C 33 a + a + 2a = 0 123 AC These are five equations in five unknowns, and may be solved by standard means. A a) The accelerations a and a may be eliminated by using 12 312A12 The tension T may be eliminated by using A

AC33 Combining and solving for a gives 3 – 4m m + m m + m m 3 4m m + m m + m m 12 23 13 b) The acceleration of the pulley B has the same magnitude as a and is in the 3

TTm c) a = g – A = g – C = g – 3 (g – a ). 13 m 2m 2m 111 Substituting the above expression for a gives 3

4m m – 3m m + m m 1 4m m + m m + m m 12 23 13 d) A similar analysis (or, interchanging the labels 1 and 2) gives 4m m – 3m m + m m 4m m + m m + m m 2 12 23 13 e) & f) Once the accelerations are known, the tensions may be found by substitution into the appropriate equation of motion, giving 4m m m 8m m m A 4m m + m m + m m C 4m m + m m + m m 12 23 13 12 23 13 g) If m = m = m and m = 2m, all of the accelerations are zero, T = 2mg and 12 3 C T = mg. All masses and pulleys are in equilibrium, and the tensions are equal to the A weights they support, which is what is expected.

a) F 2 = 62 N, which is insufficient to raise either block; a = a = 0. 12 b) F 2 = 62 N. The larger block (of weight 196 N) will not move, so a = 0, but the 1

smaller block, of weight 98 N, has a net upward force of 49 N applied to it, and so will 2 10.0 kg c) F 2 = 212 N, so the net upward force on block A is 16 N and that on block B is 1 20.0 kg 2 10.0 kg

5.127: Before the horizontal string is cut, the ball is in equilibrium, and the vertical component of the tension force must balance the weight, so T cos ? = w, or A

T = w cos ?. At point B, the ball is not in equilibrium; its speed is instantaneously 0, so A

6.2: a) ÔÇ£Pulling slowlyÔÇØ can be taken to mean that the bucket rises at constant speed, so the tension in the rope may be taken to be the bucket’s weight. In pulling a given length of rope, from Eq. (6.1), b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq. (6.2) gives the negative of the result of part (a), or – 265 J . c) The net work done on the bucket is zero.

6.4: a) The friction force to be overcome is f=?Án=?Ámg=(0.25)(30.0kg)(9.80m/s2)=73.5N, kk

b) From Eq. (6.1), Fs = (73.5 N)(4.5 m) = 331 J . The work is positive, since the worker is pushing in the same direction as the crate’s motion.

c) Since f and s are oppositely directed, Eq. (6.2) gives d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero.

?Á mg (0.25)(30kg)(9.80m/s2) F = k = = 99.2 N, cos ? – ?Á sin ? cos 30?? – (0.25) sin 30?? k

keeping extra figures. b) Fs cos? = (99.2 N)(4.50 m) cos 30?? = 386.5 J , again keeping an extra figure. c) The normal force is mg + F sin ? , and so the work done by friction is -(4.50m)(0.25)((30kg)(9.80m/s2)+(99.2N)sin30??)=-386.5J.d)Boththenormal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero.

6.6: From Eq. (6.2), 6.7:2Fscos?=2(1.80?ù106N)(0.75?ù103m)cos14??=2.62?ù109J,or2.6?ù109Jto two places.

6.8: The work you do is: rr F ? s = ((30N) ^ – (40N) ^) ? ((-9.0m)i^ – (3.0m) ^) ij j = (30 N)(-9.0 m) + (-40 N)(-3.0 m) = -270 N ? m + 120 N ? m = -150 J

6.9: a) (i) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is – mg ( y – y ). When y = y , W = 0 . 2 1 1 2 mg (ii)Letlbethelengthofthestring.W=-mg(y-y)=-mg(2l)=-25.1J mg 2 1 The displacement is upward and the gravity force is downward, so it does negative work.

1 ? ? 1 m/s ??2 2 ? ? 3.6 km / h ?? b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the speed of any object increases the kinetic energy by a factor of four.

K=1 )((4km/hr)1m/s)2=4.32?ù103J.Theperson’s 6.11: For the T-Rex, (7000 kg 2 3.6 km/hr

6.12: (a) Estimate: v ? 1m/s (walking) v ? 2 m/s (running) m ? 70 kg Walking:KE=1mv2=1(70kg)(1m/s)2=35J 22 Running:KE=1(70kg)(2m/s)2=140J 2

(b) Estimate: v ? 60 mph = 88 ft / s ? 30 m / s m ? 2000 kg KE=1(2000kg)(30m/s)2=9?ù105J 2

(c) KE = W = mgh Estimate h ? 2 m gravity KE=(1kg)(9.8m/s2)(2m)?20J

6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier. W =K -K tot 2 1

1 K = mv 2 , K = 0 1 02 2 tot mg f W = -mg(y – y ) = -mgh mg 2 1 W=-fs=-(?Ámgcos?)(h/sin?)=-?Ámgh/tan? fkk Substituting these expressions into the work-energy theorem and solving for v gives 0

W = KE 11 – mgh = mv2 – mv2 f0 22 v = v2 + 2gh 0f = (25.0m/s)2+2(9.80m/s2)(15.0m) = 30.3 m / s (b) W = KE 11 – mgh = mv2 – mv2 f0 22 v2 – v2 (30.3 m / s)2 – 02 h= 0 f = 2g 2(9.80ms/s2) = 46.8 m

6.15: a) parallel to incline: force component = mg sin? , down incline; displacement =h/sin?,downincline W=(mgsin?)(h/sin?)=mghperpendiculartoincline:nodisplacementinthis ||

mg || b) W = K – K gives mgh = 1 mv2 and v = 2gh , same as if had been dropped tot 2 1 2 from height h. The work done by gravity depends only on the vertical displacement of the When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction. When the slope angle is large, the force component in the direction of the displacement along the incline is larger but c) h = 15.0 m , so v = 2gh = 17.1s

6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the work done by friction, by a factor of four. With the stopping force given as being independent of speed, the distance must also increase by a factor of four.

6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so W = (1/ 2)mv2 = (1/ 2)(0.145 kg)(32.0 m/s)2 = 74.2 J.

so(1/2)mv2=(1/2)mv2,or,withm=2m,v2=2v2.a)Solvingfortheratioofthe AA BB B AA B speeds, v / v = 2 . b) The boats are said to start from rest, so the elapsed time is the AB distance divided by the average speed. The ratio of the average speeds is the same as the ratio of the final speeds, so the ratio of the elapsed times is t / t = v / v = 2 . BAAB

21 1 kinetic energies depend on the magnitudes of velocities only.

6.20: From Equations (6.1), (6.5) and (6.6), and solving for F,

K 1 m(v 2 – v 2 ) 1 (8.00 kg)((6.00 m / s) 2 – (4.00 m / s) 2 ) F = = 2 2 1 = 2 = 32.0 N. s s (2.50 m)

K 1 (0.420 kg)((6.00 m / s)2 – (2.00 m / s)2 ) 6.21: s = = 2 = 16.8 cm F (40.0 N)

W 2(-28.4 J) b) v = v2 + 2 = (25.0 m / s)2 + = 15.26 m / s . 21 m (0.145 kg) c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up. On the way down, gravity will have done both negative and positive work on the ball, but the net work will be the same.

6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1) gives W=Fs=mgs=(4.80kg)(9.80m/s2)(25.0m)=1176J.

b) Since the melon is released from rest, K = 0 , and Eq. (6.6) gives 1

2 6.25: a) Combining Equations (6.5) and (6.6) and solving for v algebraically, 2

W 2(10.0 N)(3.0 m) v = v 2 + 2 tot = (4.00 m / s) 2 + = 4.96 m / s. 21 m (7.00 kg)

Keeping extra figures in the intermediate calculations, the acceleration is a=(10.0kg?m/s2)/(7.00kg)=1.429m/s2.FromEq.(2.13),withappropriatechangein notation, v 2 = v 2 + 2as = (4.00 m / s) 2 + 2(1.429 m / s 2 )(3.0 m), 21 giving the same result.

6.26: The normal force does no work. The work-energy theorem, along with Eq. (6.5), gives 2K 2W ? v = = = 2gh = 2gL sin , mm whereh=Lsin?istheverticaldistancetheblockhasdropped,and?istheanglethe plane makes with the horizontal. Using the given numbers,

k work done is – ?Á mgs . The change in kinetic energy is K = -K = -(1/ 2) mv2 , and so k 10 s = v 2 / 2?Á g . b) From the result of part (a), the stopping distance is proportional to the 0k square of the initial speed, and so for an initial speed of 60 km/h, s=(91.2m)(60.0/80.0)2=51.3m.(Thismethodavoidstheintermediatecalculationof ?Á , which in this case is about 0.279.) k

6.28: The intermediate calculation of the spring constant may be avoided by using Eq. (6.9) to see that the work is proportional to the square of the extension; the work needed 3.00 cm

6.29: a) The magnitude of the force is proportional to the magnitude of the extension or (160 N)(0.015 m / 0.050 m) = 48 N, (160 N)(0.020 m / 0.050 m) = 64 N. b) There are many equivalent ways to do the necessary algebra. One way is to note ? 169 N ? that to stretch the spring the original 0.050 m requires 1 ? ? = (0.050 m) 2 = 4 J , 2 ? 0.050 m ? so that stretching 0.015 m requires (4 J)(0.015 / 0.050) 2 = 0.360 J and compressing 0.020 m requires (4 J)(0.020 / 0.050) 2 = 0.64 J . Another is to find the spring constant k=(160N)?À(0.050m)=3.20?ù103N/m,fromwhich(1/2)(3.20?ù103N/m)(0.015m)2 =0.360Jand(1/2)(3.20?ù103N/m)(0.020m)2=0.64J.

6.30: The work can be found by finding the area under the graph, being careful of the c) 1/ 2 (12 m)(10 N) = 60 J .

?6.9 F ( x) dx = ?06.9 (-20.0 N) ?06.9 N dx – 3.0 xdx 0m

N =(-20.0N)x|6.9-(3.0 )(x2/2)|6.9 00 m = -138 N ? m – 71.4 N ? m = -209.4 J or – 209 J The work is negative because the cow continues to advance as you vainly attempt to push her backward.

6.33: W = K – K tot 2 1 1 K = mv2, K =0 102 2 Work is done by the spring force. W = – 1 kx2 , where x is the amount the spring is tot 2 11 – kx2=- mv2 and x=v m/k=8.5cm 00 22 6.34: a) The average force is (80.0 J) /(0.200 m) = 400 N , and the force needed to hold the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance quadruples the work so an extra 240 J of work must be done. The maximum force is Both parts may of course be done by solving for the spring constant k=2(80.0J)?À(0.200m)2=4.00?ù103N/m,givingthesameresults.

?Á mg = kd or ?Á = = 1.76 , which is quite large. (Keeping extra figures in (20.0 N / m)(0.086 m) s s (0.100kg)(9.80m/s2) the intermediate calculation for d gives a different answer.) b) In Example 6.6, the relation ?Á11 mgd+ kd2= mv2 k1 22 was obtained, and d was found in terms of the known initial speed v . In this case, the 1

condition on d is that the static friction force at maximum extension just balances the spring force, or kd = ?Á mg . Solving for v2 and substituting, s1

k v2= d2+2gd?Ád 1k m k ? ?Á mg ?2 ? ?Á mg ? = ? s ? + 2?Á g? s ? m? k ? k ? k ? mg 2 = (?Á 2 + 2?Á ?Á ) s sk k ? (0.10 kg)(9.80 m / s 2 ) 2 ? = ? ?((0.60) 2 + 2(0.60)(0.47)), ? (20.0 N / m) ? 1

6.36: a) The spring is pushing on the block in its direction of motion, so the work is positive, and equal to the work done in compressing the spring. From either Eq. (6.9) or 22 b) The work-energy theorem gives 2W 2(0.06 J) m (4.0 kg)

6.37: The work done in any interval is the area under the curve, easily calculated when the areas are unions of triangles and rectangles. a) The area under the trapezoid is 4.0 N ? m = 4.0 J . b) No force is applied in this interval, so the work done is zero. c) The area of the triangle is 1.0 N ? m = 1.0 J , and since the curve is below the axis (F <0),theworkisnegative,or-1.0J. d)Thenetworkisthesumoftheresultsof x

done between x = 3.0 m and x = 4.0 m , so the speed is the same, 2.00 m/s. c) K = 3.0 J , so v = 2K / m = 2(3.0 J) /(2.0 kg) = 1.73 m / s .

6.39:a)Thespringdoespositiveworkonthesledandrider;(1/2)kx2=(1/2)mv2,or v = x k / m = (0.375 m) (4000 N / m) /(70 kg) = 2.83 m / s . b) The net work done by thespringis(1/2)k(x2-x2),sothefinalspeedis 12 k (4000 N / m v = ( x 2 – x 2 ) = ((0.375 m) 2 – (0.200 m) 2 ) = 2.40 m / s. 12 m (70 kg)

6.40: a) From Eq. (6.14), with dl = Rd? , P? 20 0 P0 1 r rr In an equivalent geometric treatment, when F is horizontal, F ? dl = Fdx , and the total work is F = 2w times the horizontal distance, in this case (see Fig. 6.20(a)) R sin ? , 0

w tan ? 0 0 2wR sin ? sin ? ? c) 0 = 2 0 = 2 cot 0 . wR(1 – cos ? ) (1 – cos ? ) 2 00

6.41: a) The initial and final (at the maximum distance) kinetic energy is zero, so the positive work done by the spring, (1/ 2)kx2 , must be the opposite of the negative work donebygravity,-mgLsin?,or 2mgLsin? 2(0.0900kg)(9.80m/s2)(1.80m)sin40.0?? x = = = 5.7 cm. k (640 N / m) At this point the glider is no longer in contact with the spring. b) The intermediate calculation of the initial compression can be avoided by considering that between the point 0.80 m from the launch to the maximum distance, gravity does a negative amount ofworkgivenby-(0.0900kg)(9.80m/s2)(1.80m-0.80m)sin40.0??=-0.567J,andso the kinetic energy of the glider at this point is 0.567 J. At this point the glider is no longer in contact with the spring.

on the brick by the spring and gravity is zero, so (1 2)kd 2 – mgh = 0 , or d = 2mgh / k = 2(1.80 kg)(9.80 m / s2 )(3.6 m) /(450 N / m) = 0.53 m. The spring will provide an upward force while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length.

6.43:Energy=(power)(time)=(100W)(3600s)=3.6?ù105J 1 2 6.44: Set time to stop: ?F = ma : ?Á mg = ma k

a = ?Á g = (0.200)(9.80 m / s2 ) = 1.96 m / s2 k v = v + at 0

0=8.00m/s-(1.96m/s2)t t = 4.08 s KE 1 mv2 P= =2 tt 1 (20.0 kg)(8.00 m / s 2 ) = 2 = 157 W 4.08 s

6.45:Thetotalpoweris(165N)(9.00m/s)=1.485?ù103W,sothepowerperrideris 742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long periods).

P = (700 kg) (9.8 m s2 ) (2.5 m s) = 17.15 kW. So, 17.15 kW 75 kW. = 0.23, or about 23% of the engine power is used in climbing.

6.48: a) The number per minute would be the average power divided by the work (mgh) required to lift one box, (0.50 hp) (746 W hp) = 1.41 s, (30 kg) (9.80 m s2 ) (0.90 m) or 84.6 min. b) Similarly, (100 W) = 0.378 s, (30kg)(9.80m s2)(0.90m) or 22.7 min.

6.49: The total mass that can be raised is (40.0 hp) (746 W hp)(16.0 s) = 2436 kg, (9.80 m s2 ) (20.0 m) so the maximum number of passengers is 65.0 kg

6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19), Wh (3800 N) (2.80 m) t (4.00 s)

(0.70) P (0.70) (280,000 hp)(746 W hp) 6.51: F = ave = = 8.1?ù106 N. v (65 km h) ((1 km h) (3.6 m s))

6.52: Here, Eq. (6.19) is the most direct. Gravity is doing negative work, so the rope r must do positive work to lift the skiers. The force F is gravity, and F = mg, where is the number of skiers on the rope. The power is then P=( mg)(v)cos? ? 1m s ? =(50)(70kg)(9.80ms2)(12.0kmh)? ?cos(90.0??-15.0??) ? 3.6 km h ? Note that Eq. (1.18) uses ? as the angle between the force and velocity vectors; in this case, the force is vertical, but the angle 15.0?? is measured from the horizontal, so ? = 90.0?? -15.0?? is used.

speed is v = at and the force is ma, so the power is P = Fv = (ma) (at) = ma2t. b) The power at a given time is proportional to the square of the acceleration, tripling the acceleration would mean increasing the power by a factor of nine. c) If the magnitude of the net force is the same, the acceleration will be the same, and the needed power is proportional to the time. At t = 15.0 s , the needed power is three times that at 5.0 s, or 108 W.

6.54: dK d ? 1 ? = ? mv2 ? dt dt ? 2 ? dv = mv dt = mva = mav = Fv = P.

Let M = total mass and T = time for one revolution KE = ? 1 (dm)v2 2 M dm = dx L 2?x v= T 1 ? M ? ? 2?x ?2 L

KE = ? ? dx ? ? ? 2? L ?? T ? 0 ? ?? 2?L 1 M 4? ? ? x2dx = ? ? ? 2?L??T2?0 1 ? M ? ? 4? 2 ? ? L3 ? 2 = ? ? ? ? ? ? = ? 2 ML2 T 2 2?L??T2??3? 3 5 revolutions in 3 seconds ? T = 3 5 s 2 3

6.57: a) (140 N) (3.80 m) = 532 J b) (20.0 kg) (9.80 m s2 ) (3.80 m) (- sin 25??) = -315 J d) W = – f s = – ?Á ns = – ?Á mgs cos ? fkkk

= -(0.30) (20.0 kg) (9.80 m s2 ) (3.80 m) cos 25?? = -203 J f) The result of part (e) is the kinetic energy at the top of the ramp, so the speed is v=2Km=2(14.7J)(20.0kg)=1.21ms.

d) If both the man and the child can do work at the rate of 70 J kg, and if the child only needs to use 1.96 J kg instead of 3.92 J kg, the child should be able to do more pull ups.

6.59: a) Moving a distance L along the ramp, s = L, s = L sin ?, so IMA = 1 . in out sin ? b) If AMA = IMA, (F F ) = (s s ) and so (F ) (s ) = (F ) (s ) , or W = W . out in in out out out in in out in c)

d) W (F )(s ) F F AMA W (F )(s ) s s IMA in in in in out

w -W s (7.35?ù103 J) g g (9.80m s2)(18.0m) W 8.25?ù103 J s 18.0 m c) The weight is mg = Wg = 408 N, so the acceleration is the net force divided by the s

458 N – 408 N 41.7 kg 6.61: a) 1 1 ? 2?R ?2 ? 2?(6.66 ?ù106 m) ?2 1 mv2 = m? ? = (86,400 kg) ? ? = 2.59 ?ù1012 J. 2 2 ? T ? 2 ? (90.1 min) (60 s min) ? b) (1 2) mv2 = (1 2) (86,400 kg) ((1.00 m) (3.00 s))2 = 4.80 ?ù103 J.

W = – f s = – ?Á mg cos ?s fkk = -(0.31) (5.00 kg) (9.80 m s2 ) cos 12.0??(1.50 m) = -22.3 J (keeping an extra figure) b) (5.00 kg) (9.80 m s2 ) sin 12.0?? (1.50 m) = 15.3 J. e) K = K + W = (1 2) (5.00 kg) (2.2 m s)2 – 7.0 J = 5.1 J, and so 21 2

6.63: See Problem 6.62: The work done is negative, and is proportional to the distance s thatthepackageslidesalongtheramp,W=mg(sin?-?Ácos?)s.Settingthisequalto k

the (negative) change in kinetic energy and solving for s gives (1/ 2)mv2 v2 s=- 1 = 1 mg(sin? – ?Á cos?) 2g(sin? – ?Á cos?) kk (2.2 m/s)2 2(9.80 m / s2 )(sin12?? – (0.31) cos12??) As a check of the result of Problem 6.62, (2.2 m / s) 1 – (1.5 m) /(2.6 m) = 1.4 m / s .

6.64: a) From Eq. (6.7), x x dx ? 1 ? x2 ? 1 1 ? ? x ? 2 ? ? ? ? 22

? x1 ? 2 1 ? 11x?xxx xx The force is given to be attractive, so F < 0 , and k must be positive. If x > x , 1 < 1 , x 2 1x x 21

and W < 0 . b) Taking ÔÇ£slowlyÔÇØ to be constant speed, the net force on the object is zero, so the force applied by the hand is opposite F , and the work done is negative of that found in part (a), or k ( 1 - 1 ) , which is positivx e if x > x . c) The answers have the same xx 21 12

?Á = 0.100 + Ax k when x = 12.5 m , ?Á = 0.600 k A=0.500/12.5m=0.0400/m

(a) W = KE :W = KE – KE ffi – ? ?Á mgdx = 0 – 1 mv2 ki 2 g?0x (0.00+ =1v2 f Ax)dx i 2 ? x2 ? 1 g?(0.100)x + A f ? = v2 fi ? 2? 2 ? x2 ? 1 (9.80m/s2)(0.100)x+(0.0400/m)f?=(4.50m/s)2 ?f ? 2? 2 Solve for x : x = 5.11m ff (b) ?Á = 0.100 + (0.0400 / m)(5.1 m) = 0.304 k

(c) W = KE – KE ffi 1 – ?Á mgx = 0 – mv2 k1 2 (4.50 m / s)2 x=v2/2?Ág= =10.3m i k 2(0.100)(9.80m/s2)

a b) ?x3 = (4.00 N m3)(2.00 m)3 = 32.0 N. c) Equation 6.7 gives the work needed to b

move an object against the force; the work done by the force is the negative of this, x? 2

1 W = ?0x2 Fdx = ?x k b c 2 – 2 + 3 = x2 – x3 + x4 (kx bx cx )dx 222 0 234 =(50.0N/m)x2-(233N/m2)x3+(3000N/m3)x4 222

a) When x = 0.050 m , W = 0.115 J , or 0.12 J to two figures. b) When 2

x = -0.050 m, W = 0.173 J , or 0.17 J to two figures. c) It’s easier to stretch the spring; 2

the quadratic – bx2 term is always in the – x -direction, and so the needed force, and 2

6.69: a) T = ma = m v2 = (0.120kg) (0.70 m/s)2 = 0.147 N , or 0.15 N to two figures. b) At rad R (0.40 m)

the later radius and speed, the tension is (0.120kg) (2.80 m/s)2 = 9.41 N , or 9.4 N to two (0.0 m) figures. c) The surface is frictionless and horizontal, so the net work is the work done by the cord. For a massless and frictionless cord, this is the same as the work done by the person, and is equal to the change in the block’s kinetic energy, K-K=(1/2)m(v2-v2)=(1/2)(0.120kg)((2.80m/s)2-(0.70m/s)2)=0.441J.Note 21 21 that in this case, the tension cannot be perpendicular to the block’s velocity at all times; the cord is in the radial direction, and for the radius to change, the block must have some non-zero component of velocity in the radial direction.

6.70: a) This is similar to Problem 6.64, but here ? > 0 (the force is repulsive), and 21

?1 1? – – – W=?? – ?=(2.12?ù1026N?m2((0.200m1)-(1.25?ù109m1)) ?x x ? 12

Note that x is so large compared to x2 that the term 1 is negligible. Then, using Eq. 1x 1

(6.13)) and solving for v , 2 2W 2(-2.65?ù10-17 J) v = v2 + = (3.00 ?ù105 m / s)2 + = 2.41?ù105 m / s. 2 1 ?ù – 27 m (1.67 10 kg) b) With K = 0, W = -K . Using W = – ? , 21x 2

2 K mv2 (1.67 ?ù10-27 kg)(3.00 ?ù105 m / s)2 11 c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00 ?ù105 m / s .

v(t)=dx=2?t+3?t2, a(t)=2?+6?t dt b) ma = (6.00 kg)(2(0.20 m / s2 ) + 6(0.02 m / s3 )(4.00 s) = 5.28 N. 212

6.72: In Eq. (6.14), dl = dx and ? = 31.0?? is constant, and so W=?PFcos?dl=?xFcos?dx 22

P1 x1 ???1.50m 1.00 The final speed of the object is then 2W 2(3.39 J) 21 m (0.250 kg)

6.73: a) K – K = (1/ 2)m(v2 – v2 ) 21 21 b)Theworkdonebygravityis-mgh=-(80.0kg)(9.80m/s2)(5.20m)=-4.08?ù103J, sotheworkdonebytherideris-910J-(-4.08?ù103J)=3.17?ù103J.

? xn (-n -1))xn-1 (n -1)xn-1 x0 0 Note that for this part, for n > 1, x1-n ? 0 as x ? ? . b) When 0 < n < 1 , the improper integral must be used, ?b - -? W = lim ? ( x n 1 - x0 1 )?, x2???(n -1) 2 ? n-1 and because the exponent on the x is positive, the limit does not exist, and the integral 2

diverges. This is interpreted as the force F doing an infinite amount of work, even though 2

1 kx 2 = 1 mv 2 , so 202 k 400 N / m 0 m 0.0300 kg b) W must now include friction, so 1 mv2 = W = 1 kx2 – fx , where f is the magnitude tot 2 tot 2 0 0 of the friction force. Then, k 2f v = x2 – x 00 mm 400 N / m 2(6.00 N) 0.0300 kg (0.0300 kg) c) The greatest speed occurs when the acceleration (and the net force) are zero, or kx = f , x = = = 0.0150 m . To find the speed, the net work is f 6.00 N k 400 N / m

W=1k(x2-x2)-f(x-x),sothemaximumspeedis tot 2 0 0

k 2f v = (x2-x2)- (x-x) max 0 0 mm 400 N / m 2(6.00 N) = ((0.060m)2-(0.0150m)2)- (0.060m-0.0150m) (0.0300kg) (0.0300kg) = 5.20 m/s, which is larger than the result of part (b) but smaller than the result of part (a).

6.77: Denote the initial compression of the spring by x and the distance from the initial position by L. Then, the work done by the spring is 1 kx2 and the work done by friction is 2

– ?Á mg(x + L) ; this form takes into account the fact that while the spring is compressed, k

the frictional force is still present (see Problem 6.76). The initial and final kinetic energies are both zero, so the net work done is zero, and 1 kx2 = ?Á mg(x + L) . Solving for 2k L, (1/2)kx2 (1/2)(250N/m)(0.250m)2 L = – x = – (0.250 m) = 0.813 m, ?Á mg (0.30)(2.50 kg)(9.80 m / s2 ) k

6.78: The work done by gravity is W = -mgL sin ? (negative since the cat is moving g

up), and the work done by the applied force is FL, where F is the magnitude of the applied force. The total work is tot The cat’s initial kinetic energy is 1 mv2 = 1 (7.00 kg)(2.40 m / s)2 = 20.2 J , and 212

0 relations are 11 0 22 Combining to eliminate k and then x, the two inequalties are v2 mg2 5g v2 a) Using the given numbers, (20.0 m / s)2 x > = 8.16 m, 5(9.80 m / s2 ) (1700kg)(9.80m/s2)2 (20.0 m / s) 2 b) A distance of 8 m is not commonly available as space in which to stop a car.

6.80: The students do positive work, and the force that they exert makes an angle of 30.0?? with the direction of motion. Gravity does negative work, and is at an angle of 60.0?? with the chair’s motion, so the total work done is W =((600N)cos30.0??-(85.0kg)(9.80m/s2)cos60.0??)(2.50m)=257.8J,andsothe tot speed at the top of the ramp is 2W 2(257.8 J) 21 m (85.0 kg) Note that extra figures were kept in the intermediate calculation to avoid roundoff error.

6.81: a) At maximum compression, the spring (and hence the block) is not moving, so the block has no kinetic energy. Therefore, the work done by the block is equal to its initial kinetic energy, and the maximum compression is found from 1 kX 2 = 1 mv2, or 22

that done by friction (on the block on the table). The work done by gravity is (6.00 kg) gh and the work done by friction is – ?Á (8.00 kg) gh, so k

tot 1 W = (m + m )v2, tot 1 2 2 so 2(58.8 J) (14.00 kg)

6.83: See Problem 6.82. Gravity does positive work, while friction does negative work. Setting the net (negative) work equal to the (negative) change in kinetic energy, 1 (m – ?Á m )gh = – (m + m )v2, 1 k2 1 2 2 and solving for ?Á gives k

m + (1 2) (m + m )v2 gh ?Á=1 12 k m 2

(6.00 kg) + (1 2) (14.00 kg) (0.900 m s)2 ((9.80 m s2 ) (2.00 m)) = (8.00 kg) = 0.79.

6.84: The arrow will acquire the energy that was used in drawing the bow (i.e., the work done by the archer), which will be the area under the curve that represents the force as a function of distance. One possible way of estimating this work is to approximate the F vs. x curve as a parabola which goes to zero at x = 0 and x = x , and has a maximum of F 00 at x = 0 , so that F(x) = 0 x(x – x). This may seem like a crude approximation to the x 4F 2 x02 0 x0 4F x0 4F ? x2 x3 ? 2 02000 x2 x2 ? 2 3 ? 3 0 00 0 WithF=200Nandx=0.75m,W=100J.Thespeedofthearrowisthen 00

2W=2(100J)=89ms.OtherwaysoffindingtheareaunderthecurveinFig.(6.28) m (0.025 kg) should give similar results.

k f tot W =K -K tot 2 1 K=1mv2,K=1mv2=1m(0.45v2)=0.2025(1mv2) 1202222 0 20 Thework-energyrelationgives-(0.25mg)s=(0.2025-1)1mv2 20 The mass divides out and solving gives s = 1.5 m.

6.86: Your friend’s average acceleration is v – v 6.00 m/s a = 0 = = 2.00 m/s2 t 3.00 s Since there are no other horizontal forces acting, the force you exert on her is given by F = ma = (65.0 kg)(2.00 m/s2 ) = 130 N net

Her average velocity during your pull is 3.00 m/s, and the distance she travels is thus 9.00 m. The work you do is Fx = (130 N)(9.00 m) = 1170 J , and the average power is therefore 1170 J/3.00 s = 390 W. The work can also be calculated as the change in the kinetic energy.

1.10?ù105J+1.30?ù105J 60 s 6.88: P = Fv = mav =m(2?+6?t)(2?t+3?t2) =m(4?2t+18??t2+18?2t3) =(0.96N/s)t+(0.43N/s2)t2+(0.043N/s3)t3.

6.89: Let t equal the number of seconds she walks every day. Then, (280J/s)t+(100J/s)(86400s-t)=1.1?ù107J.Solvingfort,t=13,111s=3.6hours.

b) The steady output of the athlete is 500 W/70 kg = 7 W/kg, which is below the 10 W/kg necessary to stay aloft. Though the athlete can expend 1400 W/70 kg = 20 W/kg for short periods of time, no human-powered aircraft could stay aloft for very long. Movies of early attempts at human-powered flight bear out this observation.

6.91: From the chain rule, P = d W = d (mgh) = dm gh, for ideal efficiency. Expressing dt dt dt the mass rate in terms of the volume rate and solving gives

(2000 ?ù106 W) m3 (0.92)(9.80 m/s2 )(170 m)(1000 kg/m3 ) s

2m dv d 2Pt 2P d 2P 1 P b) a = = = t = = . . dt dt m m dt m 2 t 2mt

? 2P ? 1 2P 2 8P 2 3 a) x – x = v dt = t 2 dt = t 3 = t 2 . 2 0 m m 3 9m 3

car, 13.4?ù106 W = 177, (2.8?ù103 N)(27 m/s) b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav. To climb a hill of angle ? , the extra power needed is Mg sin ?v. These will be nearly the same if a ~ g sin ? ; if g sin ? ~ g tan ? ~ 0.10 m/s2 , the power is c)(1.10?ù106kg)(9.80m/s2)(0.010)(27m/s)=2.9MW.d)Thepowerpercarneeded is that used in part (a), plus that found in part (c) with M being the mass of a single car. The total number of cars is then 13.4 ?ù106 W – 2.9 ?ù106 W = 36, (2.8?ù103N+(8.2?ù104kg)(9.80m/s2)(0.010))(27m/s) rounding to the nearest integer.

0 1 c) Approximating sin ? , by tan ? , and using the component of gravity down the incline as mg sin ? , 2

6.96: a) Along this path, y is constant, and the displacement is parallel to the force, so 2

b) Since the force has no y-component, no work is done moving in the y-direction. c) Along this path, y varies with position along the path, given by y = 1.5x, so F=?(1.5x)x=1.5?x2,and x

P = Fv = (F + F )v roll air = ((0.0045)(62.0 kg)(9.80 m/s2 ) +(1/2)(1.00)(0.463m2)(1.2kg/m3)(12.0m/s)2)(12.0m/s) = 513 W.

P 28.0 ?ù103 W 6.98: a) F = = =1.68?ù103 N. v (60.0 km/h)((1 m/s)/(3.6 km/h))

b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 + 0.35 / 4), and so the power at the lower speed is (28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp.

(8.00 hp)(746 W/hp) 6.99: a) = 358 N. (60.0 km/h)((1 m/s)/(3.6 km/h))

b) The extra power needed is 60.6 km/h mgv = (1800 kg)(9.80 m/s2) sin(arctan(1/10)) = 29.3 kW = 39.2 hp, || 3.6 km/h m/s

so the total power is 47.2 hp. (Note: If the sine of the angle is approximated by the tangent, the third place will be different.) c) Similarly, 60.0 km/h mgv = (1800 kg)(9.80 m/s2 ) sin(arctan(0.010)) = 2.94 kW = 3.94 hp, || 3.6 km/h m/s

This is the rate at which work is done on the car by gravity. The engine must do work on the car at a rate of 4.06 hp. d) In this case, approximating the sine of the slope by the tangent is appropriate, and the grade is (8.00 hp)(746 W/hp) = 0.0203, (1800 kg)(9.80 m/s2 )(60.0 km/h)((1 m/s)/(3.6 km/h))

x 0 2 0 Then, x ?2? 0 ?2?

1 ? Ax2 ? 0 2?2? To eliminate x, note that the box comes to a rest when the force of static friction balances the component of the weight directed down the plane. So, mg sin ? = Ax mg cos ?; solve sin ? A cos ? Then,

? ? sin ? ?2 ? ? A? ? ? 1v2=+g??sin?+?Acos??cos??, 2 0 ?sin A cos ? 2 ? ?? ?? ??

l = L is the moving end of the spring. Then the velocity of the point corresponding to l, denoted u, is u(l) = v 1 (when the spring is moving, l will be a function of time, and so u L

is an implicit function of time). The mass of a piece of length dl is dm = M dl, and so L

1 1 Mv2 dK = dmu2 = l2dl, 2 2 L3 and Mv2 L Mv2 2L3 6 0

22 c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball and the spring, so 1 kx2 = 1 mv2 + 1 Mv2. Solving for v, 226

0 the engine does in moving the plane forward against the resisting force. In terms of the range R and the (presumed) constant speed v, ? ?? 0 ? v2 ? In terms of the time of flight T , R = vt, so ? ?? 0 ? v? a) Rather than solve for R as a function of v, differentiate the first of these relations with respect to v, setting 0 = 0 to obtain dR F + R dF = 0. For the maximum range, dR = 0, so dW dv dv dv dv dF = 0. Performing the differentiation, dF = 2?v – 2 ? v3 = 0, which is solved for dv dv

???14 ?3.5?ù105N?m2s2?14 ? ? ? ? 0.30 N ? s2 m2 ? b) Similarly, the maximum time is found by setting d (Fv) = 0; performing the dv differentiation, 3?v2 – ? v2 = 0, which is solved for ? ? ?1 4 ? 3.5 ?ù105 N ? m2 s2 ?1 4 ? 3? ? ? 3(0.30 N ? s2 m2 ?

6.103: a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen consumption rate appears to be about 12 cm3 kg ? min, and so the total energy is

b) The run will take 6 min. Using an estimation of the rate from the graph of about 33cm3kg?mingivesanenergyconsumptionofabout2.8?ù105J.c)Theruntakes4 min, and with an estimated rate of about 50 cm3 kg ? min, the energy used is about 2.8?ù105 J. d) Walking is the most efficient way to go. In general, the point where the slope of the line from the origin to the point on the graph is the smallest is the most efficient speed; about 5 km h.

rr 6.104: From F = ma, F = ma , F = ma and F = ma . The generalization of Eq. (6.11) x xy y z z is then dv dv dv xxyyzz dx dy dz The total work is then ?(x ,y ,z ) W = F dx + F dy + F dz 222 tot x y z ( x1 , y1 , z1 )

? ? x dv ? y dv ? z dv ? = m? 2 v x dx + 2 v y dy + 2 v z dz ? xyz ? x1 dx y1 dy z1 dz ? = ? ?vx + ?v + ?vz ? m? x 2 y 2 z 2 ? v dv v dv v dv ? 1 v y1 y 2 ? xx y zz

7.2: a) For constant speed, the net force is zero, so the required force is the sack’s weight, (5.00 kg)(9.80 m s2 ) = 49 N. b) The lifting force acts in the same direction as the sack’s motion, so the work is equal to the weight times the distance, (49.00 N) (15.0 m) = 735 J; this work becomes potential energy. Note that the result is independent of the speed, and that an extra figure was kept in part (b) to avoid roundoff error.

7.3: In Eq. (7.7), taking K = 0 (as in Example 6.4) and U = 0, K = U + W . 1 2 2 1 other Friction does negative work – fy, so K = mgy – fy; solving for the speed v , 22

2(mg-f)y 2((200kg)(9.80ms2)-60N)(3.00m) v = = = 7.55 m s . 2 m (200 kg)

7.4: a) The rope makes an angle of arcsin(3.0 m ) = 30?? with the vertical. The needed 6.0 m horizontalforceisthenwtan?=(120kg)(9.80ms2)tan30??=679N,or6.8?ù102Nto two figures. b) In moving the bag, the rope does no work, so the worker does an amount of work equal to the change in potential energy, (120kg)(9.80ms2)(6.0m)(1-cos30??)=0.95?ù103J.Notethatthisisnottheproduct of the result of part (a) and the horizontal displacement; the force needed to keep the bag in equilibrium varies as the angle is changed.

7.5: a) In the absence of air resistance, Eq. (7.5) is applicable. With y – y = 22.0 m, 12 solving for v gives 2

2121 b) The result of part (a), and any application of Eq. (7.5), depends only on the magnitude of the velocities, not the directions, so the speed is again 24.0 m s. c) The ball thrown upward would be in the air for a longer time and would be slowed more by air resistance.

K = 0, W = -(35 N) ?ù (2.5 m) = -87.5 J, and taking U = 0 and 2 other 1

U=mgy=(12kg)(9.80ms2)(2.5msin30??)=147J,v=2(147J+87.5J)=6.25ms,or 2 2 1 12 kg 6.3 m s to two figures. Or, the work done by friction and the change in potential energy are both proportional to the distance the crate moves up the ramp, and so the initial speed is proportional to the square root of the distance up the ramp; (5.0 m s) 2.5 m = 6.25 m s. 1.6 m b) In part a), we calculated W and U . Using Eq. (7.7), other 2 K=1(12kg)(11.0ms)2-87.5J-147J=491.5J 22

2 m (12 kg) 7.7: As in Example 7.7, K = 0, U = 94 J, and U = 0. The work done by friction is 22 3

3 3 12 Kg 7.8: The speed is v and the kinetic energy is 4K. The work done by friction is proportional to the normal force, and hence the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same speed but four times the mass, the kinetic energy is quadrupled.

7.9: In Eq. (7.7), K = 0, W is given as – 0.22 J, and taking 1 other U = 0, K = mgR – 0.22 J, so 22

free-fall with acceleration 9.8 m s2 downward. The maximum height it reaches is therefore (v2 – v2 ) 2(-g) = 9.0 cm. The distance it travels in the first 1.25 ms can be y 0y (b) 1 W = KE = mv2 2

= 1 ?ù -6 2 (210 10 g) (130 cm s) 2 = 1.8 ergs = 1.8 ?ù10-7 J

K +U +W = K +U 1 1 other 2 2 U =0,U =mg(2R)=28,224J,W =W 1 2 other f

1 K = mv2 = 37,500 J, K = 1 mv2 = 3840 J 1 1 222 2 f221

7.12: Tarzan is lower than his original height by a distance l(cos 30 – cos 45), so his speed is v= 2gl(cos30??-cos45??)=7.9ms, a bit quick for conversation.

r motion, and so W = Fs = (110 N) (8.0 m) = 880 J. b) Because the applied force F is parallel to the ramp, the normal force is just that needed to balance the component of the weight perpendicular to the ramp, n = w cos ?, and so the friction force is f = ?Á mg cos ? and the work done by friction is kk W=-?Ámgcos?s=-(0.25)(10.0kg)(9.80ms2)cos37??(8.0m)=-157J, fk keepinganextrafigure.c)mgssin?=(10.0kg)(9.80ms2)(8.0m)sin37??=472J, again keeping an extra figure. d) 880 J – 472 J – 157 J = 251 J. e) In the direction up the ramp, the net force is F – mg sin ? – ?Á mg cos ? k

=110N-(10.0kg)(9.80ms2)(sin37??+(0.25)cos37??) = 31.46 N, so the acceleration is (31.46 N) 10.0 kg) = 3.15 m s2 . The speed after moving up the rampisv=2as=2(3.15ms2)(8.0m)=7.09ms,andthekineticenergyis (1 2)mv2 = 252 J. (In the above, numerical results of specific parts may differ in the third place if extra figures are not kept in the intermediate calculations.)

7.14: a) At the top of the swing, when the kinetic energy is zero, the potential energy (withrespecttothebottomofthecirculararc)ismgl(1-cos?),wherelisthelengthof the string and ? is the angle the string makes with the vertical. At the bottom of the swing,thispotentialenergyhasbecomekineticenergy,somgl(1-cos?)=1mv2,or 2

v=2gl(1-cos?)=2(9.80ms2)(0.80m)(1-cos45??)=2.1ms.b)At45??fromthe vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radialcomponentoftheweight,ormgcos?=(0.12kg)(9.80ms2)cos45??=0.83N.c) At the bottom of the circle, the tension is the sum of the weight and the radial acceleration, mg+mv2l=mg(1+2(1-cos45??))=1.86N, 2 or 1.9 N to two figures. Note that this method does not use the intermediate calculation of v.

7.15: Of the many ways to find energy in a spring in terms of the force and the distance, one way (which avoids the intermediate calculation of the spring constant) is to note that the energy is the product of the average force and the distance compressed or extended. a) (1 2)(800 N)(0.200 m) = 80.0 J. b) The potential energy is proportional to the square of the compression or extension; (80.0 J) (0.050 m 0.200 m)2 = 5.0 J.

2 w = mg is suspended. y = mg , and k = F , where x and F are the quantities that kx ÔÇ£calibrateÔÇØ the spring. Combining, 1(mg)2 1((60.0kg)(9.80ms2))2 U = = = 36.0 J 2 F x 2 (720 N 0.150 m)

7.17: a) Solving Eq. (7.9) for x, x = 2U = 2(3.20 J) = 0.063 m. k (1600 N m) b) Denote the initial height of the book as h and the maximum compression of the spring by x. The final and initial kinetic energies are zero, and the book is initially a height x + h above the point where the spring is maximally compressed. Equating initial and final potential energies, 1 kx2 = mg(x + h). This is a quadratic in x, the solution to 2

which is mg ? 2kh ? x = ?1 ?? 1 + ? k ? mg ? (1.20kg)(9.80ms2)? 2(1600Nm)(0.80m)? = ?1 ?? 1 + ? (1600Nm) ? (1.20kg)(9.80ms2)? The second (negative) root is not unphysical, but represents an extension rather than a compression of the spring. To two figures, the compression is 0.12 m.

7.18: a) In going from rest in the slingshot’s pocket to rest at the maximum height, the potential energy stored in the rubber band is converted to gravitational potential energy; b) Because gravitational potential energy is proportional to mass, the larger pebble c) The lack of air resistance and no deformation of the rubber band are two possible assumptions.

7.19: The initial kinetic energy and the kinetic energy of the brick at its greatest height are both zero. Equating initial and final potential energies, 1 kx2 = mgh, where h is the 2

112 K = 0.0040 J, so U = 0.0210 J = 1 kx2, so x = ?? 2(0.0210 J) = ??0.092 m. In the absence 2 2 2 5.00 N m of friction, the glider will go through the equilibrium position and pass through x = -0.092 m with the same speed, on the opposite side of the equilibrium position.

7.21: a) In this situation, U = 0 when x = 0, so K = 0.0250 J and 22 v = 2(0.0250 J) = 0.500 m s. b) If v = 2.50 m s, 2 0.200 kg 2

K=(12)(0.200kg)(2.50ms)2=0.625J=U,sox=2(0.625J)=0.500m.Or, 2 1 1 5.00 N m because the speed is 5 times that of part (a), the kinetic energy is 25 times that of part (a), and the initial extension is 5 ?ù 0.100 m = 0.500 m.

7.22: a) The work done by friction is W =-?Ámgx=-(0.05)(0.200kg)(9.80ms2)(0.020m)=-0.00196J, other k

soK=0.00704Jandv=2(0.00704J)=0.27ms.b)InthiscaseW=-0.0098J,so 2 2 0.200 kg other

2 2 0.200 kg c) In this case, K = 0, U = 0, so 22 U + W = 0 = 0.0250 J – ?Á (0.200 kg) (9.80 m s2 ) ?ù (0.100 m), or ?Á = 0.13. 1 other k k

7.23: a) In this case, K = 625,000 J as before, W = -17,000 J and 1 other U = (1 2)ky 2 + mgy 222 =(12)(1.41?ù105Nm)(-1.00m)2+(2000kg)(9.80ms2)(-1.00) The kinetic energy is then K = 625,000 J – 50,900 J – 17,000 J = 557,100 J , 2

corresponding to a speed v = 23.6 m s. b) The elevator is moving down, so the friction 2 force is up (tending to stop the elevator, which is the idea). The net upward force is then -mg+f-kx=-(2000kg)(9.80ms2)+17,000N-(1.41?ù105Nm)(-1.00m)=138,400 2 for an upward acceleration of 69.2 m s .

22 Dividing the first by the second gives x = v2 , and substituting this into the second gives 5g

k = 25 mg 2 , so a) & b), 2 v (2.50 m s)2 x = = 0.128 m, 2 5(9.80 m s ) (1160 kg)(9.80 m s2)2 (2.50 m s) 2

7.25:a)Gravitydoesnegativework,-(0.75kg)(9.80ms2)(16m)=-118J.b) Gravity does 118 J of positive work. c) Zero d) Conservative; gravity does no net work on any complete round trip.

c) Gravity is conservative, as the work done to go from one point to another is path- independent.

r 7.27: a) The displacement is in the y-direction, and since F has no y-component, the b) P r r x 12 N/m 2

r 7.28: a) From (0, 0) to (0, L), x = 0 and so F = 0,, and the work is zero. From (0, L) to r r rr (L, L), F and dl are perpendicular, so F ? dl = 0. and the net work along this path is rr zero. b) From (0, 0) to (L, 0), F ? dl = 0. From (L, 0) to (L, L), the work is that found in the example, W = CL2, so the total work along the path is CL2. c) Along the diagonal 2 rr path, x = y, and so F ? dl = Cy dy ; integrating from 0 to L gives CL . 2 (It is not a 2

coincidence that this is the average to the answers to parts (a) and (b).) d) The work depends on path, and the field is not conservative.

7.29: a) When the book moves to the left, the friction force is to the right, and the work is – (1.2 N)(3.0 m) = -3.6 J. b) The friction force is now to the left, and the work is again – 3.6 J. c) – 7.2 J. d) The net work done by friction for the round trip is not zero, and friction is not a conservative force.

7.30: The friction force has magnitude ?Á mg = (0.20)(30.0 kg)(9.80 m/s2 ) = 58.8 N. a) k

For each part of the move, friction does – (58.8 N)(10.6 m) = -623 J, so the total work 7.31: The magnitude of the friction force on the book is 2 k

a) The work done during each part of the motion is the same, and the total work done is – 2(3.68 N)(8.0 m) = -59 J (rounding to two places). b) The magnitude of the displacement is 2 (8.0 m), so the work done by friction is – 2(8.0 m)(3.68 N) = -42 N. c) The work is the same both coming and going, and the total work done is the same as in part (a), – 59 J. d) The work required to go from one point to another is not path independent, and the work required for a round trip is not zero, so friction is not a conservative force.

7.32:a)1k(x2-x2)b)-1k(x2-x2).Thetotalworkiszero;thespringforceis 212212 1323132223 221

7.33: From Eq. (7.17), the force is dU d ? 1 ? 6C x 6 ? 6? 7 dx dx x x The minus sign means that the force is attractive.

x dx x 7.35: ?U = 2kx + k?y, ?U = 2ky + k?x and ?U = 0, so from Eq. (7.19), ?x ?y ?z r j

r 7.36: From Eq. (7.19), F = – ?U i^ – ?U ^j, since U has no z-dependence. ?x ?y ?U = -2? and ?U = -2? , so ?x x3 ?y y3

r ?r r13 r7 b) Setting F = 0 and solving for r gives r = (2a b)1/6. This is the minimum of r min c) ab U (r ) = – min r12 r 6 min min ab =- ((2a/b)1/6)12 ((2a/b)1/6)6 ab2 b2 b2 4a2 2a 4a To separate the particles means to remove them to zero potential energy, and requires the negativeofthis,orE=b24a.d)TheexpressionsforEandrintermsofaandb 0 0 min are b2 2a 0 min 4a b Multiplying the first by the second and solving for b gives b = 2E r 6 , and substituting 0 min this into the first and solving for a gives a = E r12 . Using the given numbers, 0 min = ?ù -18 ?ù -10 12 = ?ù -138 ? 12 a (1.54 10 J)(1.13 10 m) 6.68 10 J m b (Note: the numerical value for a might not be within the range of standard calculators, and the powers of ten may have to be handled seperately.)

7.38: a) Considering only forces in the x-direction, F = – dU , and so the force is zero x dx when the slope of the U vs x graph is zero, at points b and d. b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point.

so the tension in each is one-third of the man’s weight. The tension in the rope is the forceheexerts,or(70.0kg)(9.80ms2)3=229N.b)Themanhasrisen1.20m,andso theincreaseinhispotentialenergyis(70.0kg)(9.80ms2)(1.20m)=823J.Inmovingup a given distance, the total length of the rope between the pulleys and the platform changes by three times this distance, so the length of rope that passes through the man’s handsis3?ù1.20m=3.60m,and(229N)(3.6m)=824J.

7.40: First find the acceleration: v2 – v2 (3.00 m s)2 a = 0 = = 3.75 m s2 2(x – x ) 2(1.20 m) 0

Then, choosing motion in the direction of the more massive block as positive: F = Mg – mg = (M + m)a = Ma + ma net M (g – a) = m(g + a) M g + a (9.80 + 3.75) m s2 = = = 2.24 m g-a – 2 (9.80 3.75) m s M = 2.24 m

Since M + m = 15.0 kg : 2.24m + m = 15.0 kg m = 4.63 kg M = 15.0 kg – 4.63 kg = 10.4 kg

7.41: a) K + U + W = K + U 1 1 other 2 2 U =U =K =0 122 W = W = -?Á mgs, with s = 280 ft = 85.3 m other f k The work-energy expression gives 1 mv2 – ?Á mgs = 0 21k 1k a) 15 mph over speed limit so $150 ticket.

1 kx2 = 1 mv2, or 22 k 400 N m m 2.00 kg b) Using energy methods directly, the initial potential energy of the spring is the final gravitationalpotentialenergy,1kx2=mgLsin?,or 2

1 kx2 1 (400 N m)(0.220 m)2 mg sin? 2 ?? (2.00 kg)(9.80 m s ) sin 37.0

7.43: The initial and final kinetic energies are both zero, so the work done by the spring is the negative of the work done by friction, or 1 kx2 = ?Á mgl, where l is the distance the 2k block moves. Solving for ?Á , k

(1/ 2)kx2 (1/ 2)(100 N m)(0.20 m)2 k2 mgl (0.50 kg)(9.80 m s )(1.00 m)

7.44: Work done by friction against the crate brings it to a halt: f x = potential energy of compressed spring k

360 J f = = 64.29 N k 5.60 m The friction force working over a 2.00-m distance does work f x = (-64.29 N)(2.00 m) = -128.6 J. The kinetic energy of the crate at this point is thus k 360 J – 128.6 J = 231.4 J, and its speed is found from mv2 = 231.4 J 2 2(231.4 J) 2 v2 = = 9.256 m2 s 50.0 kg v = 3.04 m s

AB 2A top must be at least gR. Thus, 15 22 b) U -U = (2.50)Rmg = K , so AC C

C The radial acceleration is a = v2 = 49.0 m s2 . The tangential direction is down, the C rad R normal force at point C is horizontal, there is no friction, so the only downward force is tan

7.47: a) Use work-energy relation to find the kinetic energy of the wood as it enters the 1221 Now apply work-energy relation to the motion along the rough bottom: K +U +W = K +U 1 1 other 2 2 W = W = -?Á mgs, K = U = U = 0 ; K = 78.4 J other f k 2 1 2 1 k b) Friction does – 78.4 J of work.

KE + W = PE Bottom f Top 1 mv2-?Ámgcos?d=mgh 0k 2 d = h sin? 1h v2 – ?Á g cos? = gh 2 sin? 0k

1 cos 40?? (15ms)2-(0.20)(9.8ms2) h=(9.8ms2)h 2 sin 40?? h = 9.3 m (b) Compare maximum static friction force to the weight component down the plane. f = ?Á mg cos? = (0.75)(28 kg)(9.8 m s ) cos 40?? 2 ss = 158 N mg sin? = (28 kg)(9.8 m s )(sin 40??) = 176 N > f 2 s

(c) Use same procedure as (a), with h = 9.3 m PE + W = KE Top f Bottom h1 mgh – ?Á mg cos? = mv2 sin ? 2 kB

1 1 other 2 2 mgy + 1 mv2 = 1 mv2, with h = 20.0 m and v = 10.0 m s 12122 1

v = v2 + 2gh = 22.2 m s 21 b) Use K + U + W = K + U , with point 1 at B and point 2 where the spring has 1 1 other 2 2 U = U = K = 0 ; K = 1 mv2 with v = 22.2 m s 1221211 W =W+W=-?Ámgs-1kx2,withs=100m+x other f el k 2 1 other 1 mv2 – ?Á mgs – 1 kx2 = 0 21k2 Putting in the numerical values gives x2 + 29.4x – 750 = 0. The positive root to this b) When the spring is compressed x = 16.4 m the force it exerts on the stone is F = kx = 32.8 N. The maximum possible static friction force is el 2 ss The spring force is less than the maximum possible static friction force so the stone remains at rest.

1 y = gt 2 2 12 20 m = (9.8 m s )t 2 2 t = 2.0 s 40 m v t = 40 m ? v = = 20 m s Top Top 20 s Energy conservation: KE = PE + KE Bottom Top Top 11 mv2 = mgh + mv2 BT 22 v = v2 + 2gh BT

1 1 other 2 2 Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let y = 0 at point 2. y = 41.0 m. W = – 1 kx2, where x = 11.0 m is the amount 1 other 2 122 12 Now apply F = kx to the test pulls: F = kx so x = F k = 0.602 m.

7.52: For the skier to be moving at no more than 30.0 m s ; his kinetic energy at the bottom of the ramp can be no bigger than mv2 (85.0 kg)(30.0 m s)2 = = 38,250 J 22 Friction does – 4000 J of work on him during his run, which means his combined PE and KE at the top of the ramp must be no more than 38,250 J + 4000 J = 42,250 J. His KE at the top is mv2 (85.0 kg)(2.0 m s)2 = = 170 J 22 His PE at the top should thus be no more than 42,250 J – 170 J = 42,080 J, which gives a height above the bottom of the ramp of 42,080 J 42,080 J mg (85.0kg)(9.80ms2)

7.53: The net work done during the trip down the barrel is the sum of the energy stored in the spring, the (negative) work done by friction and the (negative) work done by gravity.Using1kx2=1(F2k),theperformer’skineticenergyatthetopofthebarrelis 22

1 (4400 N)2 2 K = – (40 N)(4.0 m) – (60 kg)(9.80 m s )(2.5 m) = 7.17 ?ù103 J, 2 1100 N m

0 spring force must balance the component of the weight down the ramp plus the largest value of the static friction, or kx = w sin ? + f . The work-energy theorem requires that 0

the energy stored in the spring is equal to the sum of the work done by friction, the work done by gravity and the initial kinetic energy, or 11 kx2=(wsin?-f)L+ mv2, 0 22 where L is the total length traveled down the ramp and v is the speed at the top of the ramp. With the given parameters, 1 kx2 = 248 J and kx = 1.10 ?ù103 N. Solving for k 20 0 gives k = 2440 N m .

1 2 1 other 2 2 b) Denote the upward distance from point 2 by h. The kinetic energy at point 2 and at the height h are both zero, so the energy found in part (a) is equal to the negative of the work done by gravity and friction, -(mg+f)h=-((2000kg)(9.80ms2)+17,000N)h=(36,600N)h,so h = 6.33?ù105 J = 17.3 m. c) The net work done on the elevator between the highest point of 3.66?ù104 J the rebound and the point where it next reaches the spring is (mg – f )(h – 3.00 m) = 3.72 ?ù104 J. Note that on the way down, friction does negative work.Thespeedoftheelevatoristhen2(3.72?ù104J)=6.10ms.d)Whentheelevator 2000 kg next comes to rest, the total work done by the spring, friction, and gravity must be the negative of the kinetic energy K found in part (c), or 3

1 3 33 3 3 2

(In this calculation, the value of k was recalculated to obtain better precision.) This is a quadratic in x , the positive solution to which is 3

1 x= 3 2(7.03?ù104 N m) ?ù[2.60?ù10N+(2.60?ù10N)+4(7.03?ù10Nm)(3.72?ù10J)] 3 32 4 4

= 0.746 m, correspondingtoaforceof1.05?ù105Nandastoredenergyof3.91?ù104J.Itshouldbe noted that different ways of rounding the numbers in the intermediate calculations may give different answers.

ky=f+mg=3.66?ù104N(theconditionthattheelevatorremainsatrestwhenthe spring is compressed a distance y; y will be taken as positive) and 1 mv2 + mgy – fy = 1 kx2 (the condition that the change in energy is the work 22 W = – fy ). Eliminating y in favor of k by y = 3.66?ù104 N leads to other k

1(3.66?ù104N)2 (1.70?ù104N)(3.66?ù104N) + 2k k (1.96?ù104 N)(3.66?ù104 N) k

This is actually not hard to solve for k = 919 N m , and the corresponding x is 39.8 m. This is a very weak spring constant, and would require a space below the operating range of the elevator about four floors deep, which is not reasonable. b) At the lowest point, the spring exerts an upward force of magnitude f + mg . Just before the elevator stops, however, the friction force is also directed upward, so the net force is ( f + mg) + f – mg = 2 f , and the upward acceleration is 2 f = 17.0 m s2 . m

7.58: One mass rises while the other falls, so the net loss of potential energy is

This is the sum of the kinetic energies of the animals. If the animals are equidistant from the center, they have the same speed, so the kinetic energy of the combination is 1 m v2 , 2 tot and 2(1.176 J) (0.7000 kg)

7.59: a) The kinetic energy of the potato is the work done by gravity (or the potential 2

?1 ? (K+U)-(K+U)=m?(v2-v2)+gy? 2 2 1 1 ?2 2 1 2? ?(12)((18.6ms)2-(30.0ms)2 ? = (0.145 kg)? ? ? – (40.0 m s)2 ) + (9.80 m s 2? ? )(53.6 m) ? = -80.0 J.

b) Similarly, ? (1 2)((11.9 m s)2 + (-28.7 m s)2 ? (K+U)-(K+U)=(0.145kg)? ? 3 3 2 2 ? – (18.6 m s)2 ) – 2 ? ? (9.80 m s )(53.6 m) ? = -31.3 J.

c) The ball is moving slower on the way down, and does not go as far (in the x-direction), and so the work done by the air is smaller in magnitude.

7.61: a) For a friction force f, the total work done sliding down the pole is mgd – fd . This is given as being equal to mgh, and solving for f gives (d – h) ? h ? d ? d? When h = d , f = 0 , as expected, and when h = 0, f = mg ; there is no net force on the fireman. b) (75 kg)(9.80 m s2 )(1 – ) = 441 N . c) The net work done is 1.0 m 2.5 m (mg – f )(d – y) , and this must be equal to 1 mv2 . Using the above expression for f, 2

1 mv2 = (mg – f )(d – y) 2 ?h? = mg? ?(d – y) ?d? ? y? = mgh?1 – ?, ? d? When y = d , v = 0 ; the fireman is at the top of the pole.

at the top minus the work done by friction, K = mgh – W = (60.0 kg)(9.8 N kg)(65.0 m) – 10.500 J, or 1F

1 1 m 60 kg b) K + K – (W + W ) = 27,720 J – (?Á mgd + f d ), K = 27,720 J – [(.2)(588 N) ?ù 2 1 F A k air 2 (82 m) + (160 N)(82 m)], , or K = 27,720 J – 22,763 J = 4957 J . Then, 2

2K 2(4957 J) 2 m 60 kg c) Use the Work-Energy Theorem to find the force. W = KE, F = KE d = (4957 J) (2.5 m) = 1983 N ? 2000 N.

7.63: The skier is subject to both gravity and a normal force; it is the normal force that causes her to go in a circle, and when she leaves the hill, the normal force vanishes. The vanishing of the normal force is the condition that determines when she will leave the hill. As the normal force approaches zero, the necessary (inward) radial force is the radial componentofgravity,ormv2R=mgcos?,whereRistheradiusofthesnowball.The speed is found from conservation of energy; at an angle ? , she has descended a vertical distanceR(1-cos?),so1mv2=mgR(1-cos?),orv2=2gR(1-cos?).Usingthisin 2

?2? the previous relation gives 2(1 – cos ?) = cos ? , or ? = arccos? ? = 48.2?? . This result ?3? does not depend on the skier’s mass, the radius of the snowball, or g.

7.64: If the speed of the rock at the top is vt, then conservation of energy gives the speed v from 1 mv2 = 1 mv2 + mg(2R) , R being the radius of the circle, and so b 2b2t v2 = v2 + 4gR . The tension at the top and bottom are found from T + mg = mvt2 and bt t R

package at point B, or ?Á mgL = 1 mv2 , or k 2B

2 B k gL (9.80 m s2)(3.00 m) b) W = K -U other B A =1(0.200kg)(4.80ms)2-(0.200kg)(9.80ms2)(1.60m) 2

Equivalently, since K = K = 0, U + W + W = 0 , or A B A AB BC AB A BC

7.66: Denote the distance the truck moves up the ramp by x. K = 1 mv2 , 120 U=mgLsin?,K=0,U=mgxsin?andW =-?Ámgxcos?.From 1 2 2 other r W = (K + U ) – (K + U ) , and solving for x, other 2 2 1 1 K +mgLsin? (v2 2g)+Lsin? mg(sin?+?Á cos?) sin?+?Á cos? rr

7.67: a) Taking U (0) = 0 , 023 b) K =U -U 212 =((30.0Nm)(1.00m)2+(6.00Nm2)(1.00m)3) -((30.0Nm)(0.50m)2+(6.00Nm2)(0.50m)3) = 27.75 J, 2 0.900 kg

2122212 2

kx2 (1900 N m)(0.045 m)2 v= +2gh= +2(9.80ms2)(1.20m)=7.01ms 2 m (0.150 kg)

7.70: a) In this problem, use of algebra avoids the intermediate calculation of the spring constant k. If the original height is h and the maximum compression of the spring is d, then mg (h + d ) = 1 kd 2 . The speed needed is when the spring is compressed d , and from 22 conservation of energy, mg(h + d 2) – 1 k(d 2)2 = 1 mv2 . Substituting for k in terms of 22 h+d, ? d? mg(h+d) 1 mg?h+ ?- = mv2, ? 2? 4 2 which simplifies to ?3 1 ? ?4 4 ? Insertion of numerical values gives v = 6.14 m s . b) If the spring is compressed a distance x, 1 kx2 = mgx , or x = 2mg . Using the expression from part (a) that gives k in 2k terms of h and d, d2 d2 2mg(h+d) h+d

7.71: The first condition, that the maximum height above the release point is h, is expressed as 1 kx2 = mgh . The magnitude of the acceleration is largest when the spring is 2

compressed to a distance x; at this point the net upward force is kx – mg = ma , so the second condition is expressed as x = (m k)(g + a) . a) Substituting the second expression into the first gives 1 ? m ?2 m(g + a)2 2 ? k ? 2gh

d When the fish falls from rest, its gravitational potential energy decreases by mgy; this becomes the potential energy of the spring, which is 1 ky2 = 1 mg y2 . Equating these, 2 2d 1 mg 2d

9.73: a) a = ?2r – ?2r = (?2 – ?2 )r rad 0 0

=[?-?][?+?]r 00 =??-?0?[(?+? ] ? ? 0 )t r ?t? 0

spring is v = 2gL(sin? – ?Á cos? ) k = 2(9.80ms2)(4.00m)(sin53.1??-(0.20)cos53.1??) b) This does require energy considerations; the combined work done by gravity and frictionismg(L+d)(sin?-?Ácos?),andthepotentialenergyofthespringis1kd2, k2 where d is the maximum compression of the spring. This is a quadratic in d, which can be written as k 2mg(sin?-?Ácos?) k

Thefactormultiplyingd2is4.504m-1,anduseofthequadraticformulagives d = 1.06 m . c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has moved a distance L + d down the incline and L + d – y up the incline. The magnitude of the friction force is the same in both directions, ?Á mg cos ? , and so the work done by k

friction is – ?Á (2L + 2d – y)mg cos ? . This must be equal to the change in gravitational k potential energy, which is – mgy sin ? . Equating these and solving for y gives 2?Á cos? 2?Á sin? + ?Á cos? tan? + ?Á kk

Using the value of d found in part (b) and the given values for ?Á and ? gives k y = 1.32 m .

7.75: a)K =W -U =(20.0N)(0.25m)-(12)(40.0Nm)(.25m)2=3.75J, B other B

so v = 2(3.75 J) = 3.87 m s , or 3.9 m s to two figures. b) At this point (point C), B 0.500 kg

K = 0 , and so U = W and x = – 2(5.00 J) = -0.50 m (the minus sign denotes a C C other c 40.0 N m displacement to the left in Fig. (7.65)), which is 0.10 m from the wall.

initially stored in the spring, plus the (negative) work done by gravity and friction, or k 2

Minimizing the speed is equivalent to minimizing K ? , and differentiating the above expression with respect to ? and setting dK? = 0 gives d? 0=-mgs(cos?-?Ásin?), k

?1? or tan ? = ?Á1 , ? = arctan? ? . Pushing the box straight up (? = 90??) maximizes the ? ?Ák ? k

7.77: Let x = 0.18 m , x = 0.71 m . The spring constants (assumed identical) are then 12 known in terms of the unknown weight w, 4kx = w . The speed of the brother at a given 1 height h above the point of maximum compression is then found from 1 1?w? (4k)x2= ? ?v2+mgh, 2 2 2?g?

or (4k ) g ? x 2 ? v2 = x2 – 2gh = g? 2 – 2h?, 2 w ? x1 ? sov=(9.80ms2)((0.71m)2(0.18m)-2(0.90m))=3.13ms,or3.1mstotwo figures. b) Setting v = 0 and solving for h, 2kx2 x2 h = 2 = 2 = 1.40 m, mg 2x 1

or 1.4 m to two figures. c) No; the distance x will be different, and the ratio 1

x 0xx0 a = d 2 y dt 2 = -?2 y = -?2 y, F = ma = -m ?2 y y 00yy0 b)U=-[?Fdx+?Fdy]=m?2[?xdx+?ydy]= m?2(x2+ 12 y) xy0 0 2 c) v=dxdt=-x?sin?t=-x?(yy) x 00 0 00 0 v=dydt=+y?cos?t=+y?(xx) y 00 0 00 0 (i) When x = x and y = 0, v = 0 and v = y ? 0 x y 00 1111 K= m(v2+v2)= my2?2,U= ?2mx2andE=K+U= m?2(x2+y2) xy 00 00 000 2222 (ii) When x = 0 and y = y , v = -x ? and v = 0 0x00y 111 K= ?2mx2,U= m?2y2andE=K+U= m?2(x2+y2) 00 00 000 222 Note that the total energy is the same.

=(1000kgm3)(3.0?ù106m2)(1m)(9.8m)(150m) s2 (b) 90% of the stored energy is converted to electrical energy, so (0.90) (mgh) = 1000 kW h (0.90)?Vgh=1000kWh (1000 kW h) ( 3600 s ) V = 1h (0.90)(1000 kg m3)(150 m)(9.8 m s2) = 2.7 ?ù103 m3 Change in level of the lake: A h=V water V 2.7?ù103 m3 h= = =9.0?ù10-4m A 3.0?ù106 m2

7.81: The potential energy of a horizontal layer of thickness dy, area A, and height y is The total potential energy U is U = ?h dU = ?Ag ?0h ydy = 1 ?Ah 2 0

7.82: a) Yes; rather than considering arbitrary paths, consider that

? ?y ? 3 ??? ?? ? b) No; consider the same path as in Example 7.13 (the field is not the same). For this r rr rr force, F = 0 along Leg 1, F ? dl = 0 along legs 2 and 4, but F ? dl ? 0 along Leg 3.

rr b) Along the first leg, dy = 0 and so F ? dl = 0 . Along the second leg, x = 3.00 m , soF=-(7.50Nm2)y2,and y rr 7.83: a) Along this line, x = y , so F ? dl = -?y3dy , and y 21 y1

y 21 y1

7.84: a) r b) (1): x = 0 along this leg, so F = 0 and W = 0 . (2): Along this leg, y = 1.50 m , so rr rr F ? dl = (3.00 N m)xdx , and W = (1.50 N m)((1.50 m)2 – 0) = 3.38 J (3) F ? dl = 0 , so r W = 0 (4) y = 0 , so F = 0 and W = 0 . The work done in moving around the closed path is 3.38 J. c) The work done in moving around a closed path is not zero, and the force is not conservative.

dx potential function. For this potential, U (0) = – F 2 2k , not zero. Setting the zero of potential is equivalent to adding a constant to the potential; any additive constant will not change the derivative, and will correspond to the same force. b) At equilibrium, the force iszero;solving-kx+F=0forxgivesx=Fk.U(x)=-F2k,andthisisa 00 c)

d) No; F = 0 at only one point, and this is a stable point. e) The extreme values of x tot correspondtozerovelocity,hencezerokineticenergy,soU(x??)=E,wherex??arethe extreme points of the motion. Rather than solve a quadratic, note that k(x-Fk)-Fk,soU(x??)=Ebecomes 1 22 2

1 ? F ?2 F 2 k? x?? – ? – F k = 2 ? k? k FF x?? – = ??2 , kk FF kk f)ThemaximumkineticenergyoccurswhenU(x)isaminimum,thepointx=Fk 0

x (7.17)). b) The slope of the curve at point B is positive, so the force is negative. c) The kinetic energy is a maximum when the potential energy is a minimum, and that figures to be at around 0.75 m. d) The curve at point C looks pretty close to flat, so the force is zero. e) The object had zero kinetic energy at point A, and in order to reach a point with more potential energy than U ( A) , the kinetic energy would need to be negative. Kinetic energy is never negative, so the object can never be at any point where the potential energy is larger than U ( A) . On the graph, that looks to be at about 2.2 m. f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near 1.9 m. g) The only potential maximum, and hence the only point of unstable equilibrium, is at point C.

00 ? ? ? x2 ? ? ?? x ?2 ? x ?? x 2 x x 2 x 2 x x x02 ??? x ? ? x ??? 00 U(x)=?(1-1)=0.U(x)ispositiveforx

b) 2 ? 2? ? ? ? x ? ? x ? ? 2 m mx2 ?? x ? ? x ? ?? ? 0?? The proton moves in the positive x-direction, speeding up until it reaches a maximum speed (see part (c)), and then slows down, although it never stops. The minus sign in the square root in the expression for v(x) indicates that the particle will be found only in the 0

c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential energy. This minimum occurs when dU = 0 , or dx

dU ? ? ? x ? ? x ? ? 32 = 3 ?- 2? 0 ? + ? 0 ? ? = 0, dx x ? ? x ? ? x ? ?? 0? which has the solution x = 2x . U (2x ) = – ? , so v = ? . d) The maximum speed 0 0 4×2 2mx2 00

occurs at a point where dU = 0 , and from Eq. (7.15), the force at this point is zero. e) x=3x,andU(3x)=-dx?;v(x)=(U(x)-U(x))=[(-?)-?(x0)2-)]= 2 2 2 2 x0 1 0 0 9 x2 m 1 m 9 x2 x2 x x 2?(x0)-(x0)2-29).Theparticleisconfinedtotheregionwhere (0)__
__

8.2: See Exercise 8.3 (a); the iceboats have the same kinetic energy, so the boat with the larger mass has the larger magnitude of momentum by a factor of (2m) (m) = 2.

1 1 m2v2 1 p2 8.3: a) K = mv2 = = . 22m2m p2 p2 b) From the result of part (a), for the same kinetic energy, 1 = 2 , so the larger mass mm 12 baseball has the greater momentum; ( p p ) = 0.040 0.145 = 0.525 . From the result bird ball of part (b), for the same momentum K m = K m , so K w = K w ; the woman, with the 11 22 11 22 smaller weight, has the larger kinetic energy. (K K ) = 450 700 = 0.643. man woman

8.4: From Eq. (8.2), p=mv=(0.420kg)(4.50ms)cos20.0??=1.78kgms xx yy

8.5: The y-component of the total momentum is 8.6:FromEq.(8.2),p=-(0.145kg)(7.00ms)=-1.015kg?ms,and y

p=(0.045kg)(9.00ms)=0.405kg?ms,sothetotalmomentumhasmagnitude x

t = ( 2.00?ù10-3 s ) 8.7:p0.0450kg)(25.0ms=563N.Theweightoftheballislessthanhalfanewton,sothe weight is not significant while the ball and club are in contact.

8.8: a) The magnitude of the velocity has changed by (45.0 m s) – (- 55.0 m s) = 100.0 m s, and so the magnitude of the change of momentum is (0.145 kg) (100.0 m s) = 14.500 kg m s, to three figures. This is also the magnitude of the impulse. b) From Eq. (8.8), the magnitude of the average applied force is 14.500 kg.m/s 3 2.00?ù10 s

8.9: a) Considering the +x-components, p=p+J=(0.16kg)(3.00ms)+(25.0N)?ù(0.05s)=1.73kg?ms,andthevelocityis 21 10.8 m s in the +x-direction. b) p = 0.48 kg ? m s + (ÔÇô12.0 N)(0.05 s) = ÔÇô0.12 2

r) c) ?ù 5 ^ = (1.10 j. d) The initial velocity of the shuttle is not known; the s j m s) ^ (1.04 10 kg. m ) (95,000 kg) change in the square of the speed is not the square of the change of the speed.

1 ?0t = 2 = ?ù 7 2- ?ù 9 2 3 J F dt (0.80 10 N s)t (2.00 10 N s )t , xx 2 2

which is 18.8 kg ? m s , and so the impulse delivered between t=0 and t=2.50?ù10-3sis(18.8kg?ms)i^.b) 2

=- (9.80ms)(2.50?ù10-3s),andtheimpulseis 2 J (0.145 kg) y

(-3.55?ù10-3kg?ms)^j c)Jx=7.52?ù103N,sotheaverageforceis t 2

rrr d) p = p + j 21 =-(0.145kg)(40.0^+5.0^)m/s+(18.8^-3.55?ù10-3^j) iji = (13.0 kg.m/s)i^ – (0.73 kg.m/s) ^j.

The velocity is the momentum divided by the mass, or (89.7 m/s) i^ – (5.0m/s) ^j.

8.12: The change in the ball’s momentum in the x-direction (taken to be positive to the right) is (0.145 kg) (-(65.0 m s) cos 30o – 50.0m/s) = -15.41 kg ? m/s, so the x- component of the average force is – 15.41 kg ? m/s = -8.81?ù103 N, 1.75?ù10-3s and the y-component of the force is

(0.145kg)(65.0 m/s) sin 30?? 3 (1.75?ù10-3s) 8.13: a) J = ?t2 Fdt = A(t 2 – t1 ) + B (t 23 – t13 ), t1 3

the same magnitude as that imparted to the puck, so the player’s speed is (0.16 kg) (20.0 m s) (75.0 kg)

8.15: a) You and the snowball now share the momentum of the snowball when thrown so your speed is = 5.68 cm s. b) The change in (0.400 kg) (10.0 m s) (70.0 kg + 0.400 kg) the snowball’s momentum is (0.400 kg) (18.0 m s) = 7.20 kg ? m s), so 7.20 kg m s 70.0 kg

8.16: a) The final momentum is (0.250 kg)(-0.120 m s) + (0.350)(0.650 m s) = 0.1975 kg ? m s, taking positive directions to the right. a) Before the collision, puck B was at rest, so all of the momentum is due to puck A’s motion, and p 0.1975 kg ? m/s A1 m 0.250 kg A

111 b) K = K – K = m v2 + m v2 – m v2 2 1 A A2 B B2 A A1 222 11 = (0.250 kg) (-0.120 m s)2 + (0.350 kg)(0.650 m s)2 22 1 – (0.250 kg)(-0.7900 m s) 2 2

momentum, divided by the defender’s mass, or m v = v – A (v – v ) B2 B1 A2 A1 m B

756 N = -5.00 m s – (1.50 m s – 13.0 m s) 900 N Positive velocities are in Gretzky’s original direction of motion, so the defender has changed direction.

11 b) K – K = m (v 2 – v 2 ) + m (v 2 – v 2 ) 2 1 A A2 A1 B B2 B1 22 1 ?(756 N)((1.50 m/s)2 – (13.0 m/s)2 ) ? = 2? ? 2(9.80 m/s ) ?? + (900 N)((4.66 m/s)2 – (-5.00 m/s)2 )?? = -6.58 kJ.

8.18: Take the direction of the bullet’s motion to be the positive direction. The total momentum of the bullet, rifle, and gas must be zero, so (0.00720 kg)(601 m/s – 1.85 m/s) + (2.80 kg)(-1.85 m/s) + p = 0, gas

and p = 0.866 kg ? m s . Note that the speed of the bullet is found by subtracting gas the speed of the rifle from the speed of the bullet relative to the rifle.

AB m AA BB B A m B

K (1/2)m v2 m v2 m b) A = A A = A A = B . K (1/2)m v2 m ((m /m )v )2 m BBBBABAA

(This result may be obtained using the result of Exercise 8.3.)

8.22: 214Podecay:214Po?4?+210X 1 Setv? : KE? = m v2 ?? 2 = 2KE? v? m?

2(1.23?ù10-12J) = =1.92?ù107 m/s 6.65?ù10-27 kg

Momentum conservation: v x m 210m xp (6.65?ù10-27 kg)(1.92?ù107 m/s) = (210)(1.67 ?ù10-27 kg) =3.65?ù105m/s 0 = m?v? – m v xx

There is no net horizontal force, so P is constant. Let object A be you and object B x

0 = -m v + m v cos 35.0?? AA BB m v cos 35.0?? v = B B = 2.11 m/s A m A

8.24: Let Rebecca’s original direction of motion be the x-direction. a) From conservation of the x-component of momentum,

(45.0 kg)(13.0 m/s) = (45.0 kg)(8.0 m/s)cos 53.1?? + (65.0 kg)v , x

So v = 5.67 m s. If Rebecca’s final motion is taken to have a positive x

y – component, then (45.0 kg)(8.0 m s) sin 53.1?? y (65.0 kg) Daniel’s final speed is

v2+v2= (5.67ms)2+(-4.43ms)2=7.20ms, xy and his direction is arctan (-4.42 ) = -38?? from the x – axis, which is 91.1?? from the 5.67 direction of Rebecca’s final motion.

111 b) K = (45.0 kg) (8.0 m s)2 + (65.0 kg) (7.195 m s)2 – (45.0) (13.0 m s)2 222 = -680 J.

Kim Ken Kim Ken

m (3.00 m s) – (2.25 m s) Kim = = 0.750, m (4.00 m s) – (3.00 m s) Ken

8.26: The original momentum is (24,000 kg)(4.00 m s) = 9.60 ?ù104 kg ? m s, the final mass is 24,000 kg + 3000 kg = 27,000 kg, and so the final speed is 9.60?ù104kg?ms 2.70?ù104 kg

8.27: Denote the final speeds as v and v and the initial speed of puck A as v , and AB 0 omit the common mass. Then, the condition for conservation of momentum is v = v cos 30.0?? + v cos 45.0o 0A B AB

The 45.0?? angle simplifies the algebra, in that sin 45.0?? = cos 45.0??, and so the v terms cancel when the equations are added, giving B

v v = 0 = 29.3m s cos 30.0?? + sin 30.0?? A

From the second equation, v = A = 20.7 m s. b) Again neglecting the common mass, v B2

K (1 2)(v2 + v2 ) (29.3 m s)2 + (20.7 m s)2 2 = A B = = 0.804, K (1 2)v2 (40.0 m s)2 10

1 1 2 2 1 2 1 m1 + m2 2

velocities to the right, v = -3.00.m s and v = 1.20 m s , so v = -1.60 m s . 12

1 b) K = (0.500 kg + 0.250 kg)(-1.60 m s)2 2 11 – (0.500kg)(-3.00ms)2- (0.250kg)(1.20ms)2 22 = -1.47 J.

8.29: For the truck, M = 6320 kg, and V = 10 m s, for the car, m = 1050 kg and v = -15 m s (the negative sign indicates a westbound direction).

a) Conservation of momentum requires (M + m)v? = MV + mv , or

(6320 kg)(10 m s) + (1050 kg)(-15m s) v? = = 6.4 m s eastbound. (6320 kg + 1050 kg) – mv – (1050 kg)(-15 m s) M 6320 kg

P is constant gives x (0.250 kg)(0.200 m s) – (0.150 kg)(0.600 s) = (0.400 kg)v 2x v=-10.0cms(v=10.0cms,north) 2x 2 K = 1 (0.250 kg)(0.200 s)2 + 1 (0.150 kg)(0.600 s)2 = 0.0320 J 12 2 K = 1 (0.400 kg)(0.100 s)2 = 0.0020 J 22 K = K – K = -0.0300 J 21

Kinetic energy is converted to thermal energy due to work done by nonconservative forces during the collision.

8.32: (a) Momentum conservation tells us that both cars have the same change in momentum, but the smaller car has a greater velocity change because it has a smaller mass.

MV=mv M v (small car) = V (large car) m 3000 kg = V = 2.5 V (large car) 1200 kg (b) The occupants of the small car experience 2.5 times the velocity change of those in the large car, so they also experience 2.5 times the acceleration. Therefore they feel 2.5 times the force, which causes whiplash and other serious injuries.

8.33: Take east to be the x-direction and north to be the y-direction (again, these choices are arbitrary). The components of the common velocity after the collision are (1400 kg) (-35.0 km h) v = = -11.67 km h x (4200 kg) (2800 kg) (-50.0 km h) y (4200 kg)

as the truck had no intial x-component of momentum, so p (m + m )v cos ? v = x = car truck car mm car car 2850 kg = (16.0 m s ) cos (90?? – 24??) 950 kg = 19.5 m s.

2850 truck 1900 8.35: The speed of the block immediately after being struck by the bullet may be found from either force or energy considerations. Either way, the distance s is related to the speed v by v2 = 2?Á gs. The speed of the bullet is then block k m +m v = block bullet 2?Á gs bullet k m bullet 1.205 kg = 2(0.20)(9.80 m s 2 )(0.230 m) 5.00?ù10-3 kg = 229 m s,

8.36: a) The final speed of the bullet-block combination is 12.0?ù10-3 kg 6.012 kg Energy is conserved after the collision, so (m + M )gy = 1 (m + M )V 2 , and 2

S1 A1 Abigail before the collision. m = 80.0 kg, m = 50.0 kg, v = 6.00 m s, v = 9.00 m s. S A S 2 A2

P is constant gives x m v = m v cos 37.0?? + m v cos 23.0?? S S1 S S 2 A A2 v = 9.67 m s (Sam) S1 P is constant gives y

m v = m v sin 37.0?? – m v sin 23.0?? A A1 S S 2 A A2 v = 2.26 m s (Abigail) A1 b) K = 1 m v2 + 1 m v2 = 4101 J 1 2 S S1 2 A A1 K = 1 m v2 + 1 m v2 = 3465 J 2 2 S S 2 2 A A2 K = K – K = -640 J 21

8.38: (a) At maximum compression of the spring, v = v = V . Momentum conservation 2 10 gives (2.00 kg)(2.00 m s) = (12.0 kg)V V = 0.333 m s 1 2 = 1 (m + m )V 2 + U Energy conservation : m v 2 0 2 10 spr 22 1 2=1 2+ (2.00 kg)(2.00 m s) (12.0 kg)(0.333 m s) U spr 22 U = 3.33 J spr

(b) The collision is elastic and Eqs. (8.24) and (8.25) may be used: v = -1.33 m s, v = +0.67 m s 2 10

right) denoted as A, (3.00)v + (1.00)v = 0.200 m s. The relative velocity has switched A2 B2 direction, so v – v = -0.600 m s. Adding these eliminates v to give A2 B2 B2 (4.00)v = -0.400 m s, or v = -0.100 m s, with the minus sign indicating a final A2 A2 velocity to the left. This may be substituted into either of the two relations to obtain v = 0.500 m s; or, the second of the above relations may be multiplied by 3.00 and B2 subtracted from the first to give (4.00)v = 2.00 m s, the same result. B2 b) P = -0.009 kg ? m s, P = 0.009 kg ? m s AB = – ?ù -4 = ?ù -4 AB Because the collision is elastic, the numbers have the same magnitude.

8.41: Algebraically, v = 20 m s. This substitution and the cancellation of common B2 factors and units allow the equations in ? and ? to be reduced to

2 = cos ? + 1.8 cos ? Solving for cos? and sin ? , squaring and adding gives

(2 – ? )2 + ( )2 = Minor algebra leads to cos ? = 1.2 , or ? = 26.57??. Substitution of this result into the first 1.8 5

8.42:a)UsingEq.(8.24),vA=1u-2u=1.b)Thekineticenergyisproportionaltothe V 1u+2u 3 square of the speed, so A = 1 K c) The magnitude of the speed is reduced by a factor of K9 1 after each collision, so after collisions, the speed is (1 ) of its original value. To find 33 , consider ?1? 1 ? ? = or ? 3 ? 59,000 3 = 59,000 ln(3) = ln(59,000) ln(59,000) ln(3) to the nearest integer. Of course, using the logarithm in any base gives the same result.

AB v-v M=m A v+v A Inthiscase,v=1.50?ù107ms,andv=-1.20?ù107ms,withtheminussignindicating A

a rebound. Then, M = m 1.50 +1.20 = 9m. Either Eq. (8.25) may be used to find 1.50 +)-1.20) v = v = 3.00 ?ù106 m s, or Eq. (8.23), which gives B5 B

8.44: From Eq. (8.28), (0.30 kg)( 0.20 m) + (0.40 kg)(0.10 m) + (0.20 kg)(-0.30 m) x = = +0.044 m, cm (0.90 kg) (0.30 kg)(0.30 m) + (0.40 kg)(-0.40 m) + (0.20 kg)(0.60 m) y = = 0.056 m. cm (0.90 kg)

8.45: Measured from the center of the sun, (1.99?ù1030kg)(0)+(1.90?ù1027kg)(7.78?ù1011m) 1.99?ù1030kg+1.90?ù1027kg The center of mass of the system lies outside the sun.

8.46: a) Measured from the rear car, the position of the center of mass is, from Eq. (8.28), (1800 kg)(40.0 m) (1200 kg + 1800 kg) c) From Eq. (8.30), (1200 kg)(12.0 m s) + (1800 kg)(20.0 m s) v = = 16.8 m s. cm (1200 kg + 1800 kg) d)(1200kg+1800kg)(16.8ms)=5.04?ù104kg?ms.

8.47: a) With x = 0 in Eq. (8.28), 1 1 2 2 cm

b) P = M v cm = (0.40 kg)(5.0 m s) i^ = (2.0 kg ? m s) i^. c) In Eq. (8.32), r rr 21

8.48: As in Example 8.15, the center of mass remains at rest, so there is zero net momentum, and the magnitudes of the speeds are related by m v = m v , or 11 22 2121

8.49: See Exercise 8.47(a); with y = 0, Eq. (8.28) gives m = m (( y / y ) – 1) = 1 1 2 2 cm (0.50 kg)((6.0 m) /(2.4 m) – 1) = 0.75 kg, so the total mass of the system is 1.25 kg.

rr cm dt cm rr cm 8.50: p = 0, so F = 0. The x -component of force is zz

dp x dt dp F = y = 0.25 N y dt 8.51: a) From Eq. (8.38), F = (1600 m s)(0.0500 kg s) = 80.0 N. b) The absence of atmosphere would not prevent the rocket from operating. The rocket could be steered by ejecting the fuel in a direction with a component perpendicular to the rocket’s velocity, and braked by ejecting in a direction parallel (as opposed to antiparallel) to the rocket’s velocity.

a) Solving Eq. (8.38) for dm , dm F dt (5.22 N)(5.0 s) v 490 m s ex

8.53: Solving for the magnitude of dm in Eq. (8.39), dm ma (6000kg)(25.0m s2) v (2000 m s) ex

8.54: Solving Eq. (8.34) for v and taking the magnitude to find the exhaust speed, ex v = a m = (15.0 m s2 ) (160 s) = 2.4 km s. In this form, the quantity m is ex dm dt dm dt mt m

8.55: a) The average thrust is the impulse divided by the time, so the ratio of the average thrust to the maximum thrust is ? = 0.442. b) Using the average force in Eq. (10.0 N s) (13.3 N) (1.70 s) (8.38), v = F dt = 10.0 N?s = 800 m s. c) Using the result of part (b) in Eq. (8.40), ex dm 0.0125 kg v = (800 m s) ln (0.0258 0.0133) = 530 m s .

8.56: Solving Eq. (8.4) for the ratio m0 , with v = 0 , m0 m ? v ? ? 8.00 km s ? m ? v ? ? 2.10 km s ? ex

8.57: Solving Eq. (8.40) for m , the fraction of the original rocket mass that is not fuel, m o

m ? v? m ?v? 0 ex b) For v = 3000 m s , exp(-(3000 m s) (2000 m s)) = 0.22.

from the heights. The impulse is the mass times the sum of the speeds, J=m(v+v)=m(2gy+2gy)=(0.040kg)2(9.80ms2(2.00m+1.60m)=0.47 12 1 2 t

rr 8.59: p = ? F dt = ( ?t 3) ^j + ( ?t + ?t 2) ^j = (8.33 N s t )i^ + (30.0 Nt + 2.5 N st 2 ) ^j 3 2 23

r After0.500s,p=(1.04kg?ms)i^+(15.63kg?ms)^j,andthevelocityis

rr xx yy (-1.14N.s) b) v =v +J m=(20.0ms)+ =0.05ms (0.560N)(9.80ms2) 2 x 1x x

(0.33 N.s) 2y 1y y ((0.560 N) (9.80m s2))

8.61: The total momentum of the final combination is the same as the initial momentum; for the speed to be one-fifth of the original speed, the mass must be five times the original mass, or 15 cars.

the same magnitude that sum to zero, and hence must form the sides of an equilateral triangle. One puck will move 60?? north of east and the other will move 60?? south of east.

8.64: a) m v + m v + m v = m v , therefore A Ax B Bx C Cx tot x

(0.100kg)(0.50ms)-(0.020kg)(-1.50ms)-(0.030kg)(-0.50ms)cos60?? v= Cx 0.050kg v = 1.75 m s Cx

Similarly, (0.100kg)(0 m s) – (0.020kg)(0 m s) – (0.030 kg)(-0.50 m s)sin 60?? v= Cy 0.050 kg v = 0.26 m s Cy

b) K = 1 (0.100 kg)(0.5 m s)2 – 1 (0.020 kg)(1.50 m s)2 – 1 (0.030 kg)(-0.50 m s)2 222 – 1 (0.050 kg) ?ù [(1.75 m s)2 + (0.26 m s)2 ] = -0.092 J 2

8.65: a) To throw the mass sideways, a sideways force must be exerted on the mass, and hence a sideways force is exerted on the car. The car is given to remain on track, so some other force (the tracks on the car) act to give a net horizontal force of zero on the car, which continues at 5.00 m s east.

b) If the mass is thrown with backward with a speed of 5.00 m s relative to the initial motion of the car, the mass is at rest relative to the ground, and has zero momentum. The (200 kg ) = speed of the car is then (5.00 m s) ( ) 5.71 m s, and the car is still moving east. 175 kg

c) The combined momentum of the mass and car must be same before and after the (200 kg )(5.00 m s )+(25.0 kg )(-6.00 m s ) mass hits the car, so the speed is ( ) = 3.78 m s, with the car still 225 kg moving east.

is the same as the speed of the car, so there is no change in the velocity of either the car or the sand (the sand acquires a downward velocity after it leaves the car, and is stopped on the tracks after it leaves the car). Another way of regarding the situation is that v in ex Equations (8.37), (8.38) and (8.39) is zero, and the car does not accelerate. In any event, the speed of the car remains constant at 15.0 m/s. In Exercise 8.24, the rain is given as falling vertically, so its velocity relative to the car as it hits the car is not zero.

8.67: a) The ratio of the kinetic energy of the Nash to that of the Packard is m v = (840 kg )(9 m s ) = 22 1.68. b) The ratio of the momentum of the Nash to that of the mPvP2 (1620kg)(5ms) 2

Packard is = (840 kg)(9 m/s) = 0.933, therefore the Packard has the greater magnitude mv mPvP (1620 kg)(5 m/s) of momentum. c) The force necessary to stop an object with momentum P in time t is F = – P / t. Since the Packard has the greater momentum, it will require the greater force to stop it. The ratio is the same since the time is the same, therefore F / F = 0.933. d) P

By the work-kinetic energy theorem, F = k . Therefore, since the Nash has the greater d

kinetic energy, it will require the greater force to stop it in a given distance. Since the distance is the same, the ratio of the forces is the same as that of the kinetic energies, P

the answer in terms of the parameters, and avoids intermediate calculations, including that of the spring constant.) Let the mass of the frame be M and the mass putty be m. Denote the distance that the frame streteches the spring by x0 , the height above the frame from which the putty is dropped as h , and the maximum distance the frame moves from its initial position (with the frame attached) as d.

The collision between the putty and the frame is completely inelastic, and the common speed after the collision is v = 2gh m . After the collision, energy is 0 m+M conserved, so that 11 (m+M)v2+(m+M)gd= k((d+x)2-x2,or 0 00 22 1 m2 1 mg (2gh) + (m + M )gd = ((d + x )2 – x2 , + 00 2mM 2x 0

where the above expression for v , and k = mg x have been used. In this form, it is seen 00 that a factor of g cancels from all terms. After performing the algebra, the quadratic for d becomes ? m ? m2 d2-d?2x ?-2hx =0, ? M? 0m+M 0

which has as its positive root h ? m2 ?? ?2 d = x0 ? + ? ? ?? M ? ? M ? x ? M (m + M ) ? ?? ?0 For this situation, m = 4/3 M and h/x0 = 6, so d = 0.232 m.

speed V of the block is found from 1 M V 2 = 1 kX 2 , or V = k X . The spring constant k 2 total 2 m is determined from the calibration; k = – = 300 N m. Combining, 0.75 N 2.50 ?ù10 3 m

300 N/m (15.0 – ) 1.00 kg M 1.00 Kg (2.60 ) m 8.0?ù10-3 Kg

8.71: a) Take the original direction of the bullet’s motion to be the x-direction, and the direction of recoil to be the y-direction. The components of the stone’s velocity after impact are then

?6.00?ù10-3Kg? v = ? ?(350 m s) = 21.0 m s , x ? 0.100 Kg ?

?6.00?ù10-3Kg??(250 v =-? ms)=15.0ms, y ? 0.100 Kg ? and the stone’s speed is (21.0 m s) + (15.0 m s) = 25.8 m s , at an angle of arctan 22

(15.0)=35.5??.b)K=1(6.00?ù10-3kg)(350m/s)=368J 2 21.0 1 2 K=1(6.00?ù10-3kg)(250m/s)+1(0.100kg)(25.8m2/s2)=221J,sothecollisionis 2 22 2 not perfectly elastic.

0s after the collision is m 80.0 kg (9.9 m/s) m + m 0s 0.100 kg sv

b) Momentum is not conserved during the slide. From the work-energy theorem, the distance x is found from 1 m v2 = ?Á m gx, or 2 total k total

v 2 (5.28 m / s)2 2?Á 2 kg 8.73: Let v be the speed of the mass released at the rim just before it strikes the second Conservation of energy says 1 mv2 = mgR; v = 2gR 2

This is speed v for the collision. Let v be the speed of the combined object just after 12 the collision. Conservation of momentum applied to the collision gives mv = 2mv so v = v 2 = gR 2 1221

Apply conservation of energy to the motion of the combined object after the collision. 3

1 (2m)v2 = (2m)gy 223 v2 1 ? gR ? y3 = 2 = ? ? = R / 4 2g 2g ? 2 ? Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the initial gravitational potential energy.

mv = mv + (3m)v 013 (1) v = v + 3v 013 Energy Conservation:

1 1 1 (3m)v mv 2 = mv 2 + 2 013 2 2 2 (2) v 2 = v 2 + 3v 2 013

Solve (1) and (2) for v : v = 2.50 m s 33 Energy conservation after collision:

1(3m)v2=(3m)gh=(3m)gl(1-cos?) 3 2 Solve for ? : ? = 68.8?? 8.75: First consider the motion after the collision. The combined object has mass rr m = 25.0 kg. Apply ?F = ma to the object at the top of the circular loop, where the tot 3

v2 T + mg = m 3 R The minimum speed v for the object not to fall out of the circle is given by setting 3

3 Next, use conservation of energy with point 2 at the bottom of the loop and point 3 at the top of the loop. Take y = 0 at the point 2. Only gravity does work, so

K +U = K +U 2233 1 m v2 = 1 m v2 + m g(2R) 2 tot 2 2 tot 3 tot Use v = Rg and solve for v : v = 5gR = 13.1 m s 3 22 Now apply conservation of momentum to the collision between the dart and the 1

v2 T-mg=m 8 88 R 1600N-(8.00kg)(9.80ms2)=(8.00kg) v2 8 1.35 m v = 16.0 m s 8

mv=mv+mv 20 22 88 (2.00 kg)v = (2.00 kg)v + (8.00 kg)(16.0 m s) 02 v = v + 64.0 m s (1) 02 111 m v2 = m v2 + m v2 20 22 88 222 (2.00kg)v2=(2.00kg)v2+(8.00kg)(16.0ms) 2 02 v2 = v2 +1024m2 s2 (2) 02

Solve (1) and (2) for v : v = 40.0 m s 00 8.77: a) The coefficient of friction, from either force or energy consideration, is ?Á = v2 2gs, where v is the speed of the block after the bullet passes through. The speed k

of the block is determined from the momentum lost by the bullet, (4.00?ù10-3kg)(280ms)=1.12kg?ms,andsothecoefficientofkineticfrictionis

((1.12 kg ? ) )2 m s (0.80 kg) k m s2

b)1(4.00?ù10-3kg)((400ms)2-(120ms)2)=291J.c)Fromthecalculationofthe 2

begun to rise; this assumes a large force applied over a short time, a situation characteristic of bullets) is

The final speed v of the bullet is then p mv – MV M v= = 0 =v – V 0 mmm 1.00 kg = 450 m/s – (0.297 m/s) = 390.6 m/s, 5.00?ù10-3 kg or 390 m/s to two figures.

8.79: a ) Using the notation of Eq. ( 8.24 ), K – K = mv2 1 2 1 – mv 02 A 22 2 ??1 – ? m – M ?2 ? = 1 mv ? ?? ? ? 2 ? m+M?? ? (m + M )2 – (m – M )2 ? = K0 ? + 2 ? ? (m M ) ? ? 4mM ? ? (m + M )2 ? 0

b) Of the many ways to do this calculation, the most direct way is to differentiate the d? M ? 0 = (4mK ) ? ?, or 0 dM ? (m + M )2 ? 1 2M 0= – (m + M )2 (m + M )3 0 = (m + M ) – 2M m = M.

c) From Eq.(8.24), with m = m = m, v = 0; the neutron has lost all of its kinetic ABA energy.

M – M 2M V = A B V and V = A A M +M 0 B M +M AB AB

The ratio of the kinetic energies of the two particles after the collision is

1 M V 2 M ? V ?2 M ? M – M ?2 (M – M ) 2 2 A A = A ? A ? = A ? A B ? = A B 1 M V 2 M ? V ? M ? 2M ? 4M M 2BBBBBAAB (M – M )2 or KE = KE A B AB 4M M AB

b) i) For M = M , KE = 0; i.e., the two objects simply exchange kinetic energies. ABA

ii) For M = 5M , AB KE (4M ) 2 4 A= B = KE 4(5M )(M ) 5 B BB

A c) We want KE (M – M )2 M 2 – 2M M + M 2 A =1= A B = A A B B KE 4M M 4M M B AB AB

which reduces to M 2 – 6M M + M 2 = 0, AABB from which, using the quadratic formula, we get the two possibilities M = 5.83 M and AB

leaves the chute to point 2 just before it lands in the cart. Take y = 0 at point 2, so y1 = 4.00 m. Only gravity does work, so K +U = K +U 1122 1 mv2 + mgy = 1 mv2 21 122

v = v2 + 2gy = 9.35 m/s 211 b) In the collision between the package and the cart momentum is conserved in the horizontal direction. (But not in the vertical direction, due to the vertical force the floor exerts on the cart.) Take +x to be to the right. Let A be the package and B be the cart.

Px is constant gives m v + m v = (m + M )v a A1x B B1x A B 2 x v = -5.00 m/s B1x v = (3.00 m/s) cos 37.0? ( The horizontal velocity of the package is constant during A1x its free-fall.) Solving for v gives v = -3.29 m/s. The cart is moving to the left at 3.29 m/s after 2x 2x the package lands in it.

8.82: Even though one of the masses is not known, the analysis of Section (8.4) leading to Eq. (8.26) is still valid, and v = 0.200 m/s + 0.050 m/s = 0.250 m/s. b) The mass red m may be found from either energy or momentum considerations. From momentum red conservation, (0.040 kg)(0.200 m/s – 0.050 m/s) red (0.250 m/s) As a check, note that

r r r? + r v 2 = (v ? + v ) ? (v v ) A A cm A cm rrrrrr = v ? ? v ? + v ? v + 2v ? ? v A A cm cm A cm rr = v? 2 + v 2 + 2v ? ? v , A cm A cm

with a similar expression for v 2 . The total kinetic energy is then B

11 K = m v2 + m v2 AA BB 22 = 1 m (v? v 2 r r ) 1 m (v r r ) 2 + 2 + v ? ? v + ? 2 + 2 + 2v ? ? vv A A cm A cm B B cm B cm 22 = 1 (m + m )v2 1 (m r? 2 m r? 2 ) + 0, v + v A B cm A A B B 22 rr 2 m v v + m v? v A A cm B B cm

The last term in brackets can be expressed as r rr 2(m v ? + m v ? ) ? v , A A B B cm and the term rrrr r mv?+mv?=mv?+mv?-(m+m)v A A B B A A B B A B cm

= and so the term in square brackets in the expression for the kinetic energy vanishes, showing the desired result. b) In any collision for which other forces may be neglected the velocity of the center of mass does not change, and the 1 Mv2 in the kinetic energy 2 cm will not change. The other terms can be zero (for a perfectly inelastic collision, which is not likely), but never negative, so the minimum possible kinetic energy is 1 Mv2 . 2 cm

8.84: a) The relative speed of approach before the collision is the relative speed at which the balls separate after the collision. Before the collision, they are approaching with relative speed 2 v , and so after the collision they are receding with speed 2 v . In the limit that the larger ball has the much larger mass, its speed after the collision will be unchanged (the limit as m >> m in Eq. (8.24)), and so the small ball will move upward AB with speed 3 v . b) With three times the speed, the ball will rebound to a height time times greater than the initial height.

andso(15.0kg)v=(120.0kg)(4.00m/s-v).Solvingforv,v= =3.56ms. (120.0 kg) (4.00 m s) (135.0 kg) b) After Jack jumps, the speed of the crate is (75.0 kg) (4.00 m s) = 2.222 m s , and the (135.0 kg) momentum of Jill and the crate is 133.3 kg ? m s . After Jill jumps, the crate has a speed v and Jill has speed 4.00 m s – v, and so 133.3 kg ? m s = (15.0 kg)v – (45.0 kg)(4.00 m s – v), and solving for v gives v = 5.22 m s. c) Repeating the calculation for part (b) with Jill jumping first gives a final speed of 4.67 m s.

directions. We can thus write M V = -M V AA BB

Since M = M – M , we have AB (M – M )V = -M V BA BB V -M A= B V M-M BB

Then for the ratio of the kinetic energies KE 1 M V 2 M M 2 M A=2AA=AB=B KE 1 M V 2 M (M – M )2 M B 2 BB B B A

From the two equations M KE = B KE and KE + KE = Q ABAB M A

We can solve for KE to find B ?M? B ? ? KE = Q – KE = Q 1 – A A ? MA + MB ?

p initial momentum of the neutron is zero, so to conserve momentum the electron must be e

x ee pp v=(mm)vÔÇôp epe The total kinetic energy after decay is K = 1 m v2 + 1 m v2 . Using the tot 2 e e 2 p p

e tot 2 p p p e

K p = 1 = 1 = ?ù -4 = Thus 5.44 10 0.0544 % K 1+ m m 1836 tot p e

8.88: The ratios that appear in Eq. ( 8.42 ) are 0.0176 and 1 , so the kinetic energies are 1.0176 1.0176 a) 0.0176(6.54?ù10- J)=1.13?ù10 Jandb) 1 (6.54?ù10 J)=6.43?ù10- J. 13 -14 -13 13 1.0176 1.0176 Notethattheenergiesdonotaddto6.54?ù10-13Jexactly,duetoroundoff.

8.89: The ÔÇ£missing momentumÔÇØ is ?ù -22 ? – ?ù -25 ?ù 3 = ?ù -22 ? 5.60 10 kg m s (3.50 10 kg)(1.14 10 m/s ) 1.61 10 kg m s .

Since the electron has momentum to the right, the neutrino’s momentum must be to the left.

proton, mv = mv cos? + mv cos ? A1 A2 2 0 = mv sin? – mv sin ? A2 B2

or v = v cos? + v cos ? A1 A2 B2 A2 B2

b ) After minor algebra, v2 =v2 +v2 +2v v (cos?cos?-sin?sin?) A1 A2 B2 A2 B2 A2 B2 A2 B2

c ) For a perfectly elastic collision, 111 A1 A2 B 2 A1 A2 B 2 222 Substitution into the above result gives cos(? + ?) = 0. d) The only positive angle with 2

velocities from conservation of momentum. Taking the positive direction to the left, ?- ? (800 N)(5.00 m/s) cos 30.0 (600 N)(7.00 m/s) cos 36.9 v = = 0.105 m s 1000 N

8.93: a ) From symmetry, the center of mass is on the vertical axis, a distance (L 2)cos(? 2) from the apex. b) The center of mass is on the (vertical) axis of symmetry, a distance 2(L / 2) / 3 = L 3 from the center of the bottom of the . c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal, and each is equal to (L 2) / 2 = L 4. The distance of this point from the corner is (1 8)L=(0.353)L.Thismayalsobefoundfromconsiderationofthesituationofpart (a), with ? = 45?? d ) By symmetry, the center of mass is in the center of the equilateral triangle, a distance (L 2)(tan 60??) = L 12 = (0.289)L above the center of the base.

8.94: The trick here is to notice that the final configuration is the same as if the canoe ( assumed symmetrical ) has been rotated about its center of mass. Intially, the center of mass is a distance = 0.643 m from the center of the canoe, so in rotating about (45.0 kg) (1.5 m) (105 kg) this point the center of the canoe would move 2 ?ù 0.643 m = 1.29 m.

8.95: Neglecting friction, the total momentum is zero, and your speed will be one-fifth of the slab’s speed, or 0.40 m / s.

original parabolic trajectory, ÔÇ£landingÔÇØ at the position of the original range of the projectile. Since the explosion takes place at the highest point of the trajectory, and one fragment is given to have zero speed after the explosion, neither fragment has a vertical component of velocity immediately after the explosion, and the second fragment has twice the velocity the projectile had before the explosion. a) The fragments land at positions symmetric about the original target point. Since one lands at 1 R, the other 2

lands at 2 2 2 g 0 2 (9.80 m / s2 ) b) In terms of the mass m of the original fragment and the speed v before the explosion, K = 1 mv2and K = 1 m (2v)2 = mv2,so K = mv2 – 1 mv2 = 1 mv2. The speed v is related 1 2 2 22 2 2 to v by v = v cos ? , so 000

is at its maximum height and has zero kinetic energy. Let A be the piece with mass 1.40 kg and B be the piece with mass 0.28 kg. Let v and v be the speeds of the two pieces AB immediately after the collision.

1 m v2 + 1 m v2 = 860 J 2AA2BB Since the two fragments reach the ground at the same time, their velocitues just after the explosion must be horizontal. The initial momentum of the shell before the explosion is zero, so after the explosion the pieces must be moving in opposite horizontal AA BB

Use this to eliminate v in the first equation and solve for v : AB 2BBBA B ABAB

b) Use the vertical motion from the maximum height to the ground to find the time it takes the pieces to fall to the ground after the explosion. Take + y downward. 0y y 0

v = v sin ? – gt 90 0 = (150 m s)sin ? – 2 55.0 9.8 m s t t = 12.5 s.

The heavier fragment travels back to its starting point, so it reversed its velocity. v = v cos ? = (150 m s) cos 55? = 86.0 m s to the left after the explosion; this is v . Now x0 9 3

Mv = m v + m v 0 33 99 (12kg)(86.0ms)=(3.00kg)v+(9.00kg)(-86.0ms) 3

v = 602 m s 3 x = x + x = (86.0 m s)(12.5 s) + (602 m s)(12.5) 3 Before explosion After explosion x = 8600m from where it was launched 3

neutron that moves in the + y – direction by the subscript 1 and the emitted neutron that moves in the ÔÇôy-direction by the subscript 2. Using conservation of momentum in the x- and y-directions, neglecting the common factor of the mass of neutron, v = (2v 3) cos 10?? + v cos 45?? + v cos 30?? 0012 012

With sin 45?? = cos 45?? , these two relations may be subtracted to eliminate v and 1, rearrangement gives v (1 – (2 3) cos 10?? + (2 3)sin 10??) = v (cos 30?? + sin 30??), 02

from which v = 1.01?ù103 m s or 1.0 ?ù103 m s to two figures. Substitution of this into 2

either of the momentum relations gives v = 221 m s. All that is known is that there is no 1

z – component of momentum, and so only the ratio of the speeds can be determined. The Kr Ba

8.100: a) With block B initially at rest, v = A v . b) Since there is no net external m cm mA+mB A1 force, the center of mass moves with constant velocity, and so a frame that moves with the center of mass is an inertial reference frame. c) The velocities have only x – components, and the x -components are u = v – v = B v ,u = -v = – A u . Then, P = m u + m u = 0 . mm mA+mB mA+mB A1 A1 cm A1 B1 cm A1 cm A A1 B B1 d) Since there is zero momentum in the center-of-mass frame before the collision, there can be no momentum after the collision; the momentum of each block after the collision must be reversed in direction. The only way to conserve kinetic energy is if the

(8.27), the relative speeds are the same, and ?= 1. c) Neglecting air resistance, the speeds before and after the collision are 2gh and 2gH , and ?= 2gH1 = H h . d) 1 2 gh 1

Frompart(c),H1=?h=(0.85 m= e)Hk+1=Hk?,andbyinduction 2 ) (1.2 ) 0.87 m. 2 2

n 8.102: a) The decrease in potential energy (- < 0) means that the kinetic energy increases. In the center of mass frame of two hydrogen atoms, the net momentum is necessarily zero and after the atoms combine and have a common velocity, that velocity must have zero magnitude, a situation precluded by the necessarily positive kinetic energy. b) The initial momentum is zero before the collision, and must be zero after the collision. Denote the common initial speed as v , the final speed of the hydrogen atom as 0

v , the final speed of the hydrogen molecule as V , the common mass of the hydrogen atoms as m and the mass of the hydrogen molecules as 2 m . After the collision, the two particles must be moving in opposite directions, and so to conserve momentum, v = 2V . From conservation of energy,

1 (2m)V 1 2 – + mv2 = 3 1 mv2 20 22 mV 2 – + 2mV 2 = 3 mv2 20 v2 V2=0+ , 2 3m

from which V =1.203?ù104 m/s, or 1.20 ?ù 104 m/s to two figures and the hydrogen atom speed is v = 2.41?ù104 m/s.

dv dm m = – v – mg, ex dt dt where the positive direction is taken upwards (usually a sign of good planning). b) Diving by a factor of the mass m,

dv v dm dt m dt c)20ms2-9.80ms2=10.2ms2.d)3327ms-(9.80m/s2)(90)=2.45kms,which is about three-fourths the speed found in Example 8.17.

ex 3,300 kg ex ex ex c) (1.18)v + v In(1000 / 300) = (2.38)v . d) Setting the result of part (c) equal to ex ex ex ex ex

8.106: a) There are two contribution to F,F=v|dmdt|-v|dmdt|,orF=(v-v)dmdt net net ex net ex b) F / | dm / dt |= (1300 N) (150 kg s) = 8.66 m s net ex .

v(t)=(2400ms)ln(1(1-(t/120s)))Att=90s,thisspeedis3.33km/s,andthisisalso the speed for t > 90 s.

b) The acceleration is zero for t < 0 and t > 90 s. For 0 ? t ? 90 s, Eq. (8.39) gives, with = -m 120 s, a = ( ( 2 dt 0 1- t / 120 s ))

c) The maximum acceleration occurs at the latest time of firing, t = 90 s, at which time the acceleration is, from the result of part (a), ( ) = 80 m/s , and so the astronaut is 20 m/s2 2 1-90 / 120 subject to a force of 6.0 kN, about eight times her weight on earth.

k1 cake and v is its speed after the impulse is applied. The distance d that the cake moves duringthistimeisthend=1?Ágt2.Whileslidingonthetable,thecakemustloseits 2 k1 kineticenergytofriction,or?Á mg(r-d)=1mv2.Simplificationandsubstitutionforv k2 2

gives r – d = 1 g ?Á2 t 2 , substituting for d in terms of t 2 gives k1 2 ?Ák 2

r = 1 gt ? ?Á ?Á 2 ? 1 gt ?Á ( + ?Á ), 2 2 ? k1 + ?Á k1 ? = 2 k1 ?Á ? k2 ? ?Ák2 k1 k 2 2 which gives t = 0.59 s.

8.109: a) Noting than dm = M dx avoids the intermediate variable ? . Then, L

1 ?0L M L cm ML2 b) In this case, the mass M may be found in terms of ? and L, specifically by using dm = ?Adx = ?Adx to find that M = ?A ? xdx = ?A L2 2 . Then,

2 ?0L 2 L3 2L cm ?AL2 ?AL2 3 3

8.110: By symmetry, x = 0 . Using plane polar coordinates leads to an easier integration, and using the Theorem of Pappus (2?y (? ) ) cm

a 2 = 4 ?a3 is easiest of all, but the cm 2 3 method of Problem 8.109 involves Cartesian coordinates.

Forthex-coordinate,dm=?ta2-x2dx,whichisanevenfunctionofx,so ?xdx=0.Forthey-coordinate,dm=?t2a2-y2dy,andtherangeofintegrationis from 0 to a, so

22 yy M Making the substitutions M = 1 ??a2t, u = a2 – y2, du = -2y, and 2

T = (?x)g , where x is the length of rope over the edge, hanging vertically. In raising the rope a distance – dx , the work done is (?g ) x (- dx)(dx is negative) The total work done is then

x2 ?gl2 0 2 32 b) The center of mass of the hanging piece is initially a distance l/8 below the top of the table, and the hanging weight is (?g )(l / 4) , so the work required to raise the rope is (?g ) (l / 4) (l / 8) = ?gl 2 / 32, as before.

8.112: a) For constant acceleration a, the downward velocity is v = at and the distance x that the drop has fallen is x = 1 at2. Substitution into the differential equation gives 2

1 1 ( )2 3 at2g= at2a+at = a2t2, 222

3 1 1 ? 9.80 m/s ?( ) 2 2 2? 3 ? c) kx = (2.00 g/m) (14.7 m) = 29.4 g.

2.50 m (14.0 cm) (128??)(? rad 180??) ? rev ? ? 2? rad ? ? 1 min ? ? min ? ? rev ? ? 60 s ? b) (35???ù?rad180??)(199rads)=3.07?ù10-3s.

d? 9.3:a)?=z=(12.0rads3)t,soatt=3.5s,?=42rads2.Theangularacceleration z dt is proportional to the time, so the average angular acceleration between any two times is thearithmeticaverageoftheangularaccelerations.b)?=(6.0rads3)t2,soat z

t = 3.5 s, ? = 73.5 rad s. The angular velocity is not linear function of time, so the z average angular velocity is not the arithmetic average or the angular velocity at the midpoint of the interval.

= d? = -2 = -1.60 3 z dt z ?(3.0 s) – ?(0) – 2.20 rad s – 5.00 rad s ?av-z = = = -2.40 rad s2., 3.0 s 3.0 s which is half as large (in magnitude) as the acceleration at t = 3.0 s.

9.5:a)?=?+3?t2=(0.400rads)+(0.036rads3)t2b)Att=0,?=?= zz z av-z 5.00s The acceleration is not constant, but increasing, so the angular velocity is larger than the average angular velocity.

zz (9.00rads3)t.a)Setting?=0resultsinaquadraticint;theonlypositivetimeatwhich z

zz z av 4.23 s

= d? = – 2 = dw = – = = b 9.7: a) ? 2bt 3ct and ? z 2b 6ct. b) Setting ? 0, t . z dt z dt z 3c

9.8: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value.

(b) The angular acceleration is ? – ? 8.00 rad s – (-6.00 rad s) ? = 0 = = 2.00 rad s2 t 7.00 s Thus it takes 3.00 seconds for the wheel to stop (? = 0) . During this time its speed is z

decreasing. For the next 4.00 s its speed is increasing from 0 rad s to + 8.00 rad s . (c) We have ? = ? + ? t + 1 ?t 2 002 = 0 + (-6.00 rad s) (7.00 s) + 1 (2.00 rad s2 ) (7.00 s)2 2

Alternatively, the average angular velocity is – 6.00 rad s + 8.00 rad s = 1.00 rad s 2 Which leads to displacement of 7.00 rad after 7.00 s.

9.9: a) ? – ? = 200 rev, ? = 500 rev min = 8.333 rev s, t = 30.0 s, ? = ? 00 ?? +?? ? – ?0 = ? 0 ? t gives ? = 5.00 rev s = 300 rpm ?2? b) Use the information in part (a) to find ? : ? = ? + ?t gives ? = -0.1111 rev s2 0

z 0z z 0z z 2

(200 rev min – 500 rev min) ?ù (1min ) rev 9.11: a) 60 s = -1.25 . (4.00 s) s2 The number of revolutions is the average angular velocity, 350 rev min, times the time interval of 0.067 min, or 23.33 rev. b) The angular velocity will decrease by another 200 rev min 60 s min 1.25 rev s 2

= ? -? ?z Rewriting Eq. (9.11) as ? – ? = t(? + 1 ? t) and substituting for t gives 0 0z 2 z

? ? – ? ?? 1 ? ? – ?0 = ? z 0 z ?? ? – ? ? ?+( ) z ?? 0 z z 0 z ? ?? 2 1 ?? +? ? = (? – ? )? z 0 z ? ? z 0z ? 2 ? 1 = (?2-?20z), 2? z

b) ? = (1 2)(1 ? )(?2 – ?2 ) = (1 2) (1 (7.00 rad)) (16.0 rad s)2 – (12.0 rad s)2 ) = z z 0z 8rad s2.

? 0 z ? z 1.50 rad s 2 0z 0 2(1.50rad) s2

?-? ?t 60.0rad (2.25rads2)(4.00s) ? = 0 – z = – = 10.5 rad s. 0z t 2 4.00s 2

9.16:FromEq.(9.7),with?=0,?=?z=140rads=23.33rads2.Theangleismost 0 z z t 6.00 s av-z

9.17: From Eq. (9.12), with ? = 0, the number of revolutions is proportional to the z square of the initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.0 rev.

has rotated for each instant in time and each of the three situations:

(a) (b) (c) t rev’s ? rev’s ? rev’s ?

0.05 0.50 180 0.03 11.3 0.44 158 0.10 1.00 360 0.13 45 0.75 270 0.15 1.50 540 0.28 101 0.94 338 0.20 2.00 720 0.50 180 1.00 360 ÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇôÔÇô The ? and ? graphs are as follows: z

(24.0 rad s) (2.00 s) + (30.0 rad s2 ) (2.00 s)2 2 = 108 rad, so the total angle is 108 rad + 432 rad = 540 rad. b) The angular velocity when the circuit breaker trips is (24.0 rad s) + (30.0 rad s2 )(2.00 s) = 84 rad s, so the average angular velocity while the wheel is slowing is 42.0 rad s, and the time to slow to a stop is 432 rad = 10.3 s, so the 42.0 rad s time when the wheel stops is 12.3 s . c) Of the many ways to find the angular acceleration, the most direct is to use the intermediate calculation of part (b) to find that 84 rad s z z 10.3 s

9.20: a) Equation (9.7) is solved for ? = ? – ? = ? – ? ? t, which gives z t, or 0z z z z-ave z 2 z rad 0 z 2 z t t2 z z

9.21: The horizontal component of velocity is r? , so the magnitude of the velocity is a) 47.1 m/s

? ? ? rad/s ??2 ? ? 30 rev/min ? ?

9.22: a) 1.25 m-s = 50.0 rad s , 1.25 m- = 21.55 rad s , or 21.6 rad s to three figures. 25.0 ?ù10 3 m 58.0 ?ù10 3

c)?=50.0rads-21.55rads= ?ù-3 2 z (74.0 min) (60 s min)

rad a 400,000 ?ù 9.80 m s 2 ?= = =1.25?ù104rads, r 2.50?ù10-2 m 1 min 60 s

9.25: a) a = 0, a = ?r = (0.600 rad s2 )(0.300 m) = 0.180 m s2 and so a = 0.180 m s2. rad tan 3 rad

The tangential acceleration is still 0.180 m s2 , and so on a = (0.180 m s2 )2 + (0.377 m s2 )2 = c)Foranangleof120??,a=0.754ms2,anda=0.775ms2,sinceais rad tan still0.180 m/s2

9.26:a)?=?+?t=0.250revs+(0.900revs2)(0.200s)=0.430revs z 0z z (note that since ? and ? are given in terms of revolutions, it’s not necessary to 0z z

convert to radians). b) ? – t = (0.340 rev s) (0.2s) = 0.068 rev . c) Here, the conversion av z to radians must be made to use Eq. (9.13), and ? 0.750 m ? ( ) ?2?

d) Combining equations (9.14) and (9.15), a = a2rad + a2 tan = (?2r)2 + (?r)2 = [((0.430 rev s ?ù 2? rad rev)2 (0.375 m))2 + ((0.900 rev s2 ?ù 2? rad ] 1

rev)(0.375 m))2 2 a (3000)(9.80 m s2 ) 9.27: r = rad = ((5000 ( ))2 = 10.7 cm, ?2 ? rad s rev min) 30 rev min

?v? rad ? ? ? a 0.500 m s b) From the result of part (a), ? rad 0.250 rad v 2.00 m s

30 rev min 2 2 = (0.831m s)2 = v r (12.7?ù10 3 m) 2

9.30:a)?=atan=-10.0ms2=-50.0rads2b)Att=3.00s,v=50.0msand r 0.200 m ? = v = 50.0 m s = 250 rad s and at t = 0, v = 50.0 m s + (-10.0 m s2 ) r 0.200 (0-3.00s)=80.0ms,so?=400rads.c)?t=(325rads)(3.00s) ave

=975rad=155rev.d)v=ar=(9.80ms2)(0.200m)=1.40ms.Thisspeedwill rad

50.0 m s -1.40 m s = 4.86 s after t = 3.00 s, or at t = 7.86 s be reached at time . (There are many 10.0 m s equivalent ways to do this calculation.)

9.31: (a) For a given radius and mass, the force is proportional to the square of the angular velocity; (640 rev min ) = 2.29 (note that conversion to rad s is not necessary for this 2 423 rev min part). b) For a given radius, the tangential speed is proportional to the angular velocity; 640 = 1.51 (again conversion of the units of angular speed is not necessary). 423c)(640revmin)(?rads)(0.470m)=15.75ms,or15.7mstothreefigures,and 30 rev min 2

rad r (0.470 m 2) 9.32: (a) v = R? T

? 7.5 rev ? ? 1 min ? ? 2? rad ? 2.00 cm s = R ? ? ? ? ? ? ? min ? ? 60 s ? ? 1 rev ? R = 2.55 cm D = 2R = 5.09 cm b) a = R? T

v 5.00 m s 9.33: The angular velocity of the rear wheel is ? = r = = 15.15 rad s. r r 0.330 m The angular velocity of the front wheel is ? = 0.600 rev s = 3.77 rad s f

Points on the chain all move at the same speed, so r ? = r ? rrff r = r (? ? ) = 2.99 cm rrfr

9.34: The distances of the masses from the axis are L , L and 3L , and so from Eq. (9.16), 44 4 the moment of inertia is ? L ?2 ? L ?2 ? 3L ?2 11 ? 4 ? ? 4 ? ? 4 ? 16

22 9.35: The moment of inertia of the cylinder is M L and that of each cap is m L , so the 12 4 12 2

9.36: Since the rod is 500 times as long as it is wide, it can be considered slender. a) From Table (9.2(a )),

12 12 b) From Table (9.2(b)), 33 c) For this slender rod, the moment of inertia about the axis is obtained by considering it as a solid cylinder, and from Table (9.2(f )),

2(0.200m)2=8.00?ù10-2m2,andthemomentofinertiais ?ù -2 2 = ?ù -2 ? 2 4(0.200 kg) (0.800 10 m ) 6.40 10 kg m . b) Each sphere is 0.200 m from the a) The two masses through which the axis passes do not contribute to the moment of 2

1 ? L ?2 9.38: (a) I = I + I = M L2 + 2m ? ? bar balls 12 bar balls ? 2 ? =1(4.00kg)(2.00m)2+2(0.500kg)(1.00m)2=2.33kg?m2 12

1 (b) I = m L2 + m L2 bar ball 3 =1(4.00kg)(2.00m)2+(0.500kg)(2.00m)2=7.33kg?m2 3

c) I = 0 because all masses are on the axis (d) I = m d 2 + 2m d 2 = M d 2 bar ball Total = (5.00 kg)(0.500 m)2 = 1.25 kg ? m2

9.39: I = I + I (d = disk, r = ring) dr = (3.00 g cm )?r2 = 23.56 kg 3 disk : m dd

1 I = m r 2 = 2.945 kg ? m2 d dd 2 ring:m=(2.00gcm3)?(r2-r2)=15.08kg (r=50.0cm,r=70.0cm) r 21 1 2

f 3 and the distance multiplied by f , and so the moment of inertia is multiplied by f3(f)2=f5.b)(2.5)(48)5=6.37?ù108.

9.41: Each of the eight spokes may be treated as a slender rod about an axis through an end, so the moment of inertia of the combination is ?m ? I = m R2 + 8 ? spoke ? R2 rim ?3? ?8? = ?(1.40 kg) + (0.20 kg)? (0.300 m) 2

?3? = 0.193 kg ? m2 9.42: a) From Eq. (9.17), with I from Table (9.2(a)), 1 1 1 rev 2? rad rev 6 K= mL2?2= (117kg)(2.08m)2(2400 ?ù )2=1.3?ù10J. 2 12 24 min 60 s min b) From mgy = K, K (1.3?ù106J) mg (117 kg)(9.80 m s 2 )

9.43: a) The units of moment of inertia are [kg][m2] and the units of ? are equivalent to [s ] and so the product 1 I? has units equivalent to [kg ? m ? s- ] = [kg ? (m s) ] , -1 2 2 2 2

which are the units of Joules. A radian is a ratio of distances and is therefore unitless. b) K = ?2I?2 1800 , when is in rev min.

2221 (K – K ) I=2 2 1 (?2 – ?2 ) 21 (-500 J) = 2 ( )2 ((520 rev min)2 – (650 rev min)2 ) ? rad s 30 rev min

9.46: The work done on the cylinder is PL, where L is the length of the rope. Combining Equations (9.17), (9.13) and the expression for I from Table (9.2(g)), 1 ? 1 ? v2 (40.0 N)(6.00 m s)2 2 g 2 g L 2(9.80 m s2 )(5.00 m)

1 9.47:Expressing?intermsof?,?2=arad.CombiningwithI=MR2,Eq.(9.17) rad R 2 1 1 (70.0 kg)(1.20 m)(3500 m s)2 becomes K = MRa = = 7.35?ù104 J. rad 22 4

9.48:a)WithI=MR2,withexpressionforvis 2gh 1+ M m b) This expression is smaller than that for the solid cylinder; more of the cylinder’s mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is larger. A larger fraction of the potential energy is converted to the kinetic energy of the cylinder, and so less is available for the falling mass.

T b) Differentiating the expression found in part (a) with respect to T,

dt dt 9.50: The center of mass has fallen half of the length of the rope, so the change in gravitational potential energy is 11 22

cm P

22 4 9.53: MR2= MR2+Md2,sod2= R2,andtheaxiscomesnearesttothecenterof 3 5 15 the sphere at a distance d = (2 15)R = (0.516)R.

9.54: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its end and perpendicular to the rod,

M ? L ?2 M p cm 12 ? 2 ? 3

9.55:?p=?+md2,so?=1?(a2+b2)+?((a)2+(b)2),whichgives cm 12 2 2

12 12 9.57: In Eq. (9.19), ? = M L2 and d = (L 2 – h), so cm 12

?1 ?L ?? 2 ?p = ? ? L2 + ? – h? ? ??12 ? 2 ? ?? ?1 1 ? = ? ? L2 + L2 – Lh + h2 ? ?12 4 ? ?1 ? = ? ? L2 – Lh + h2 , ? ?3 ? which is the same as found in Example 9.12.

9.58: The analysis is identical to that of Example 9.13, with the lower limit in the integral being zero and the upper limit being R, and the mass ? = ?L?R2. The result is 2

L x2 a) ? = ? dm = ? ?x dx = ? 2 0 L yL2 = 2 0

b) ?=?Lx2(?x)dx=?x4 4 0 L 42 0 This is larger than the moment of inertia of a uniform rod of the same mass and length, since the mass density is greater further away from the axis than nearer the axis.

L c) ? = ? (L – x)2 ?xdx 0 L =??(L2x-2Lx2+x3)dx 0 L ? x2 x3 x4 ? = ?? L2 – 2L + ? ? 2 3 4? 0

L4 =? 12 M 6 This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end.

r 9.61: a) For a clockwise rotation, ? will be out of the page. b ) The upward direction rr crossed into the radial direction is, by the right-hand rule, counterclockwise. ? and r are rr r perpendicular, so the magnitude of ? ?ù r is ?r = v. c) Geometrically, ? is r rr perpendicular to v , and so ? ?ù v has magnitude ?v = a and from the right-hand rad, rule, the upward direction crossed into the counterclockwise direction is inward, the r direction of a . Algebraically, rad r r r r (r r) a =??ùv=??ù??ùr rad r(r r) r(r r) =???r-r??? r =-?r, 2

rr where the fact that ? and r are perpendicular has been used to eliminate their dot product.

For planetary alignment, earth must go through 60?? more than Mars: ? = ? + 60?? EM w t = ? t + 60?? EM 60?? t= ? -? EM 360?? 360?? w = and w = EM 1yr 1.9yr 60?? ? 365d ? t = = 0.352 yr? ? = 128 d 360?? – 360?? ? 1yr ? 1yr 1.9 yr

9.63: a) v = 60 mph = 26.82 m s r=12in.=0.3048m v ? = = 88.0 rad s = 14.0 rev s = 840 rpm r b) same ? as in part (a) since speedometer reads same r = 15 in. = 0.381 m v = r? = (0.381m)(88.0 rad s) = 33.5 m s = 75 mph c) v = 50 mph = 22.35 m s r = 10 in. = 0.254 m v ? = = 88.0 rad s; . this is the same as for 60 mph with correct tires, so r speedometer read 60 mph.

9.64: a) For constant angular acceleration ? = ? , and so a = ?2r = 2??r. 2

d? z dt dw z dt c) An extreme of angular velocity occurs when ? = 0, which occurs at z

t = = 3.20 rad/s = 2.13 s, and at this time ?2 3? 1.50rad/s3 (3.20 rad/s2 )2 z 3(0.500 rad/s3 )

9.66: a) By successively integrating Equations (9.5) and (9.3), ? ?=?t-t2=(1.80rad/s2)t-(0.125rad/s3)t2, z 2 ?? 26 ? b) The maximum positive angular velocity occurs when ? = 0, t = , the angular z?

velocity at this time is ??? ????2 1?2 1(1.80rad/s2)2 z ? ? ? 2 ? ? ? 2 ? 2 (0.25 rad/s3 ) 2? The maximum angular displacement occurs when ? = 0, at time t = (t = 0 is an z ? inflection point, and ?(0) is not a maximum) and the angular displacement at this time is

? ? ? ?2 ? ? ?3 ? 2 ? 2 2 3 2 (1.80 rad/s2 )3 2 ? ? ? 6 ? ? ? 3 ? 2 3 (0.25 rad/s3 )2

0.300 rad r 60.0 m rad d) e)a= a2rad+a2tan= (5.40ms2)2+(3.00ms2)2=6.18ms2, and the magnitude of the force is F = ma = (1240 kg)(6.18 m s2 ) = 7.66 kN.

rad atan 3.00 9.69: a) Expressing angular frequencies in units of revolutions per minute may be accomodated by changing the units of the dynamic quantities; specifically, 2W ? = ?2 + 21 I

? 2(-4000 J ) ? ? ? rad s ?2 = (300 rev min)2 + ? ? ? ? ? 16.0 kg ? m2 ? ? 30 rev min ? = 211rev min.

the bottom of the circle. The speed, from energy considerations, is v= 2gh= 2gR(1-cos?),where?istheanglefromtheverticalatrelease,and v 2g 2(9.80m s2) ? = = (1 – cos? ) = (1 – cos 36.9??) = 1.25 rad s. R R (2.50 m) b) ? will again be 0 when the meatball again passes through the lowest point. c)a isdirectedtowardthecenter,anda =?2R,a =(1.25rads2)(2.50m)=3.93m rad rad rad rad

1.696 m s r 2.00?ù10-2 m 9.72: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade.

? ? rad s ? ? 0.208 m ? ? 30 rev min ? ? 2 ? ? ? ? rad s ??2 ? 0.208 m ? rad ? ? ? ? ? ? = b) a = ?2r = 2(3450 rev min) 5.43?ù104 m s2, ? ? 30 rev min ? ? ? 2 ? so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity.

rad 0 0

=[?-?][?+?]r 00 ??-? ?[ ] = ? 0 ? (? + ? )t r ?t? 0 0

b) From the above, a (85.0 m s2 – 25.0 m s2 ) 2 ? 2(15.0 rad) c) Similar to the derivation of part (a), 1121 0 222 d) Using the result of part (c), K (45.0 J – 20.0 J) I = = = 0.208 kg ? m2. ? ? ((2.00 m s2 ) /(0.250 m))(15.0 rad)

9.74: I = I + I wood lead 22 = m R2 + m R2 wL 53 4 m = ? V = ? ?R3 w ww w 3

diameter 0.30 m (radius 0.15 m) 11 I = mR2 = (80 kg) (0.15 m)2 = 0.9 kg ? m2 22

?1 ? ?1? ? = 2? mL2 ? = 2? ? (0.160 kg)(0.250 m)2 ?3 ? ?3? = 6.67 ?ù10-3 kg ? m2

9.77: a) ? = 90.0 rpm = 9.425 rad s 1 2K 2(10.0 ?ù106 J) K = ??2 so ? = = = 2.252 ?ù105 kg ? m2 2 ?2 ( 9.425 rad s)2 m=?V=??R2t(?=7800kgm3isthedensityofironandt=0.100misthe thickness of the flywheel) 11 ? = mR2 = ??tR4 22 R = (2I ??t)1 4 = 3.68 m; diameter = 7.36 m b) a = R?2 = 327 m s2 c

9.78:Quantitatively,fromTable(9.2),I=1mR2,?=mR2and?=2mR2?a)ObjectA A2 B C3 has the smallest moment of inertia because, of the three objects, its mass is the most concentrated near its axis. b) Conversely, object B’s mass is concentrated and farthest from its axis. c) Because ? = 2 5mR2 , the sphere would replace the disk as having the sphere smallest moment of inertia.

2?2? 2?2(0.3308)(5.97?ù1024 kg)(6.38?ù106m)2 K = = = 2.14 ?ù1029 J. T2 (86,164s)2 1 ?2?R?2 2?2(5.97?ù1024kg)(1.50?ù1011m)2 b) M ? ? = = 2.66 ?ù1033 J. 2 ? T ? (3.156 ?ù107 s)2 c) Since the Earth’s moment on inertia is less than that of a uniform sphere, more of the Earth’s mass must be concentrated near its center.

9.80: Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR. The kinetic energy is 11111? K = ??2 + mv2 = ??2 + m(?R)2 = (? + mR2) 2 ? 22222 Using ? = 1 mR2 and solving for ?, 2

Consider a small strip of width dy and a distance y ? xdy ? ? yb ?? 2 dy ? ? 2M ? dm = M ? 1 bh ? = M ? ?? ? = ? ? y dy ? 2 ? ? h ?? bh ? ? h2 ? (dm)x2 = 2Mb2 dI = 1 y3 dy 3 3h4 I=?hdI=2Mb42?hy3dy= y4 2Mb2 ? 1 h ? ? | ?= 0 3h 0 ? 0 ? I = 1 Mb2 = 2.304 kg ? 6

? = 2.00 rev s = 4.00? K = 1 I?2 = 182 J 2 The strip has area x dy and the area of the sign is

1 below the top of the triangle. The 1 bh, so the mass of the strip is 2

KE=1mv2=1(8.00kg)(5.00m/s)2=100J.Thechangeinitspotentialenergywhile 22 fallingismgh=(8.00kg)(9.8m/s2)(2.00m)=156.8J The wheel must have the ÔÇ£missingÔÇØ 56.8 J in the form of rotational KE. Since its outer rim is moving at the same speed as the falling mass, 5.00 m s : v = r? ? v 5.00m/s = = = 13.51 rad/s r 0.370m KE = 1 I? 2 ; therefore 2 2KE 2(56.8 J) I = = = 0.6224 kg ? m2 or 0.622 kg ? m2 ?2 (13 )2 .51 rad s

(b) The wheel’s mass is 280 N 9.8 m s2 = 28.6 kg. The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim. Its moment of inertia would be

I = MR2 = (28.6kg)(0.370m)2 = ? 2 3.92 kg m 9.83:a)(0.160kg)(-0.500m)(9.80ms2)=-0.784J.b)Thekineticenergyofthestick is 0.784 J, and so the angular velocity is

? 2k 2k 2(0.784 J) I ML2 3 (0.160 ) (1.00 )2 kg m 3

This result may also be found by using the algebraic form for the kinetic energy, K = MgL 2, from which ? = 3g L , giving the same result. Note that ? is independent c) v = ?L = (5.42 rad s) (1.00 m) = 5.42 m s d) 2gL = 4.43 m s ; This is 2 3 of the result of part (c).

potential energy is zero (the rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance ?R below the axle, since the length of the rope is 2?R and half this distance is the position of the center of the mass. Initially, every part of the rope is moving with speed ? R, and when the rope has unwound, and 0

the cylinder has angular speed ? , the speed of the rope is ?R (the upper end of the rope has the same tangential speed at the edge of the cylinder). From conservation of energy, using I = (1 2)MR 2 for a uniform cylinder, ?M m? ?M m? ?4 2? 0 ?4 2? Solving for ? gives (4?mg R) ? = ?2 + , (M + 2m) 0

9.85: In descending a distance d, gravity has done work m gd and friction has done B

work – ?Á m gd , and so the total kinetic energy of the system is gd (m – ?Á m ). In KA BKA terms of the speed v of the blocks, the kinetic energy is K = 1 (m + m )v 2 + 1 I? 2 = 1 (m + m + I R 2 )v 2 , AB AB 2 22 where ? = v R, and condition that the rope not slip, have been used. Setting the kinetic energy equal to the work done and solving for the speed v, 2gd(m -?Á m ) m + m + I R2 AB

the initial potential energy with respect to the center of the loop when its center is directly belowthenailisgR(1-cos?).Fromthework-energytheorem,

1 K= I?2=M?2R2=MgR(1-cos?), 2 1 9.88: a) K = I?2 2 1 ? 1 ?? 2? rad s ?2 = ? (1000kg)(0.90m)2??3000rev min?ù ? 2 ? 2 ?? 60 rev min ? = 2.00 ?ù107 J.

K 2.00?ù107J b) = = 1075 s, P 1.86 ?ù104 W ave

1 1 1 -2 +(1.60kg)(5.00?ù10-2m)2) 9.89: a) M R2 + M R2 = ((0.80kg)(2.50?ù10 m)2 11 22 222 =2.25?ù10-3kg?m2.

b) See Example 9.9. In this case, ? = v R , and so the expression for v becomes 1

2gh v= 1+(I mR2) 2(9.80 m s2 )(2.00 m) = = 3.40 m s. (1+((2.25?ù10-3kg?m2) (1.50kg)(0.025m)2)) c) The same calculation, with R instead of R gives v = 4.95 m s. This does make 21 sense, because for a given total energy, the disk combination will have a larger fraction of the kinetic energy with the string of the larger radius, and with this larger fraction, the disk combination must be moving faster.

v by h? = ? , and with the form for h given in Example 9.9, h? = h . b) Considering 2

2g 1+M 2m the system as a whole, some of the initial potential energy of the mass went into the kinetic energy of the cylinder. Considering the mass alone, the tension in the string did work on the mass, so its total energy is not conserved.

? = 1 MR2 = 1 (10.0 kg)(0.150 m)2 = 0.1125 kg ? m2 22 K = 1 I?2 so ? = 2K I = 66.67 rad s 2

11 m gh = m v2 + I ?2 box box box pulley pulley 22 1 + I ?2 cylinder cylinder 2 vv ? = Box and ? = Box pulley cylinder rr p cylinder

1 1 ? 1 ? ? v ?2 mgh=mv2+?mr2??B? 2 2?2 ??r ? B BB Pp ?p? 1 ? 1 ? ? v ?2 + ? m r 2 ? ? B ? 2 ? 2 C ? ? rC ? C

111 m gh = m v2 + m v2 + m v2 B BB PB CB 244 m gh v= B 1m+1m+1m B 2B4p4C

(3.00 kg)(9.80 m s2 )(1.50 m) = = 3.68 m s 1.50 kg + 1 (7.00 kg) 4

9.93: a) The initial moment of inertia is I = 1 MR2. The piece punched has a mass of 02 M and a moment of inertia with respect to the axis of the original disk of 16

I =(25)MR2+ML2,and P ? ? ?? ?2 ? ML2 ? ? 5 ?? L ? ?

IfR=(0.05)L,thedifferenceis(25)(0.05)2=0.001.b)(I ML2)=(m 3M),which rod rod rod

2 2+ 2 9.95: a) With respect to O, each element r in Eq. (9.17) is x y , and so i ii

=?2=?2+2=?2+?2=+I I m r m (x y ) m x m y I . Oiiiiiiiiixy iiii

b) Two perpendicular axes, both perpendicular to the washer’s axis, will have the same moment of inertia about those axes, and the perpendicular-axis theorem predicts that they will sum to the moment of inertia about the washer axis, which is M (R + R 22 ), 212

=I =M 2+ 2 xy412 12 6 0 x y x y x y 12

9.96: Each side has length a and mass M , and the moment of inertia of each side about 4

an axis perpendicular to the side and through its center is 1 M a 2 = Ma . The moment of 2

cylindrical shells of thickness dr and radius r; the cross-sectional area of such a shell is 2? r dr, and the mass of shell is dm = 2? rL? dr = 2? ? Lr 2dr. The total mass of the cylinder is then M=?dm=2?L??Rr2dr=2?L?R3 03 and the moment of inertia is o 55 b) This is less than the moment of inertia if all the mass were concentrated at the edge, as with a thin shell with I = MR 2 , and is greater than that for a uniform cylinder with 2

2 9.98: a) From Exercise 9.49, the rate of energy loss is 4? ? d? ; solving for the moment of ? 3 dt inertia I in terms of the power P,

??3 1 (5?ù1031W)(0.0331s)3 1s ? = = = 1.09 ?ù1038 kg ? m2. 4? d? dt 4?2 4.22?ù10-13s 5? 5(1.08?ù1038kg?m2) b) R = = = 9.9 ?ù103 m, about 10 km. 2? 2(1.4)(1.99 ?ù1030 kg)

c) 2?R=2?(9.9?ù103m) 6m =6.3?ù10-3c. = 1.9 ?ù10 s ? (0.0331s)

?? d) = =6.9?ù1017kgm3, V (4? 3)R3

which is much higher than the density of ordinary rock by 14 orders of magnitude, and is comparable to nuclear mass densities.

mass density is I = 2 ?R2 = 8? ?R5, and so the difference in the moments of inertia of 5 15 two spheres with the same density ? but different radii 21 21

b) A rather tedious calculation, summing the product of the densities times the difference in the cubes of the radii that bound the regions and multiplying by 4? 3, gives M = 5.97 ?ù1024 kg. c) A similar calculation, summing the product of the densities times the difference in the fifth powers of the radii that bound the regions and multiplying by 8? 15, gives I = 8.02 ?ù1022 kg ? m2 = 0.334MR2.

9.100: Following the procedure used in Example 9.14 (and using z as the coordinate alongtheverticalaxis)r(z)=zR,dm=??Rz2dzandd?=Rz4dz.Then, 2 ?? 4 h h2 2 h4

44 2 h 10 h4 0 10 The volume of a right circular cone is V = 1 ?R2h, the mass is 1 ?R2hand so 33

0 02 02 gives a quadratic in ? . The positive solution is ?(t) = 1 = ? r + 2 ?vt – r ? 2

(The negative solution would be going backwards, to values of r smaller than r .) 0

c) Differentiating, d? v ? (t ) = = , z dt r 2 + 2 ?vt 0

d? ?v2 0 The angular acceleration ? is not constant. d) r = 25.0 mm; It is crucial that ? is z0 measured in radians, so ? = (1.55 ?Ám rev) (1 rev 2? rad) = 0.247 ?Ám rad. The total angle turned in 74.0 min = 4440 s is

? 2(2.47?ù10 )( )( )? -7 1 m/rad 1.25 m/s 4440 s ?= ? ? 2.47?ù10-7m/rad? +(25.0?ù10-3m)2-25.0?ù10-3m ?? ? =1.337?ù105 rad which is 2.13 ?ù 104 rev.

10.2: ? = -(8.00 N)(5.00 m) = -40.0 N ? m, 1

? = (12.0 N)(2.00 m) sin 30?? = 12.0 N ? m, 2 where positive torques are taken counterclockwise, so the net torque is – 28.0 N ? m, with the minus sign indicating a clockwise torque, or a torque into the page.

10.3: Taking positive torques to be counterclockwise (out of the page), ? = -(0.090 m) ?ù (180.0 N) = -1.62 N ? m, ? = (0.09 m)(26.0 N) = 2.34 N ? m, ?1 = ( 2 )(0.090 m) (14.0 N) = 1.78 N ? m, so the net torque is 2.50 N ? m, with the 2

3 direction counterclockwise (out of the page). Note that for ? the applied force is 3

10.4: ? + ? = -F R + F R = (F – F )R 121221 10.5: a)

A ? =0 B ? = (50 N)(sin 30??)(0.2 m) = 5 N ? m, CW C ? = (50 N)(0.2 m) = 10 N ? m, CW D

(b) ??=8.7N?m-5N?m-10N?m = -6.3 N ? m, CW 10.7: I = 2 MR 2 + 2mR 2 , where M = 8.40 kg, m = 2.00 kg 3

I = 0.600 kg ? m2 0 0 ?? = I?, ? = I? = -0.0524 N ? m f 60 rev min t (8.00 s)

1 1 ? 2? rad s ?2 2 2 ? 60 rev min ?

10.9:v= 2as= 2(0.36ms2)(2.0m)=1.2ms,thesameasthatfoundin Example 9-8.

? FR (40.0 N)(0.250 m) 10.10: ? = = = (5.0 kg ? m2 ) = 2.00 rad s2. II

? m ? ? M + 3m ? 10.11: a) n = Mg + T = g ?M + ? = g ? ? ? 1 + 2m M ? ?1 + 2m M ? b) This is less than the total weight; the suspended mass is accelerating down, so the tension is less than mg. c) As long as the cable remains taut, the velocity of the mass does not affect the acceleration, and the tension and normal force are unchanged.

a horizontal force to the right, and gravity exerts a downward force, so the normal force must exert a force up and to the left, as shown in Fig. (10.9).

b)n=(9.0N)2+((50kg)(9.80ms2))490N,atanangleofarctan(9.0)=1.1??fromthe 2 490 vertical (the weight is much larger than the applied force F ).

f ? R I? MR(? t ) 10.13: ?Á = = = = 0 k n n Rn 2n

(50.0kg)(0.260m)(850revmin)(?rads) = 30 rev min = sN

10.14: (a) Falling stone: g = 1 at 2 2 = 1 a(3.00 s)2 12.6 m 2

a = 2.80 m s2 Stone : ? F = ma : mg – T = ma(1) Pulley : ? ? = I? : TR = 1 MR2? = 1 MR2 ( a ) 22R T = 1 Ma(2) 2

=1mR2=1(8.25kg)(0.0750m)2=0.02320kg?m2 10.15: I 22 00 ?2 = ?2 + 2?(? – ? ) gives ? = -8.046 rad/s2 00 ? ? = I? ? ? = ? = – f R = – ?Á nR fkk I? – ?Á nR = I? so n = = 7.47 N k ?ÁR k

10.16: This is the same situtation as in Example 10.3. a) T = mg (1 + 2m M ) = 42.0 N. b) v = 2 gh (1 + M 2m = 11.8 m s. c) There are many ways to find the time of fall. Rather than make the intermediate calculation of the acceleration, the time is the distance divided by the average speed, or h (v 2) = 1.69 s. d) The normal force in Fig. (10.10(b)) is the sum of the tension found in part (a) and the weight of the windlass, a total 159.6 N (keeping extra figures in part ( a)).

10.17: See Example 10.4. In this case, the moment of inertia I is unknown, so a=(mg)(m+m+(IR2)).a)a=2(1.20m)(0.80s)2=3.75m/s2, 1212 1 1 11 2 2 1 b) The torque on the pulley is (T – T )R = 0.803 N ? m, and the angular acceleration is 21 1

? Fl 3F I 1 Ml 2 Ml 3 10.19: The acceleration of the mass is related to the tension by Ma = Mg – T , and the cm angular acceleration is related to the torque by I? = ? = TR, or a = T / M , where ? = a / RSt and I = MR2 have been used. cm cm a) Solving these for T gives T = Mg / 2 = 0.882 N. b) Substituting the expression for T into either of the above relations gives a = g / 2, from which cm t = 2h a = 4h g = 0.553 s. c) ? = v R = a t R = 33.9 rad/s. cm cm cm

(1 2)I ? 2 1 1 (12)Mv2+c(m12)I?2=1+(MI)v2/?2=1+MR2, cm cm cm cm Icm where v = R? for an object that is rolling without slipping has been used. cm a)I =(12)MR2,sotheaboveratiois13.b)I=(25)MR2,sotheaboveratiois cm 2 7. c) I = 2 3 MR 2 , so the ratio is 2 5. d) I = 5 8 MR 2 , so the ratio is 5 13.

10.22: a) The acceleration down the slope is a = g sin ? – f , the torque about the M

center of the shell is a2a2 ? = Rf = I? = I = MR2 = MRa, R3R3 sof=2a.Solvingtheserelationsaforfandsimultaneouslygives5a=gsin?,or M3 3 33 a= gsin?= (9.80ms2)sin38.0??=3.62ms2, 55 22 33 The normal force is Mg cos ? , and since f ? ?Á n, s

f 2 Ma 2 a 2 3 g sin ? 2 s n Mg cos ? 3 g cos ? 3 g cos ? 5

b) a = 3.62 m s 2 since it does not depend on the mass. The frictional force, however, is twice as large, 9.65 N, since it does depend on the mass. The minimum value of ?Á also s

n = mg cos ? mg sin ? – ?Á mg cos ? = ma s g(sin ? – ?Á cos ?) = a (eq.1) s

??=? =?Ámgcos?R;I=2mR2 fs 5

? ? = I? gives ?Á mg cos ? = 2 mR2? s5 No slipping means ? = a R , so ?Á g cos? = 2 a (eq.2) s5 We have two equations in the two unknowns a and ?Á . Solving gives s

a = 5 g sin ? and ?Á = 2 tan ? = 2 tan65.0?? = 0.613 7 s7 7 3

a = 3 g sin ? and ?Á = 2 tan ? = 2 tan65.0?? = 0.858 5 s5 5

The value of ?Á calculated in part (a) is not large enough to prevent slipping for the s

v = R? for no slipping cm a) Get v at bottom: 1212 mgh = mv + I? 22 1 2 1 ? 2 ? ? v ?2 mgh = mv + ? mR2 ? ? ? 2 2?5 ??R? 10 v = gh 7

Now use energy conservation. Rotational KE does not change 1 mv2 + KE = mgh? + KE Rot Rot 2 v2 10 gh 5 h? = = 7 = h 2g 2g 7 (b) mgh = mgh? ? h? = h With friction on both halves, all the PE gets converted back to PE. With one smooth side, some of the PE remains as rotational KE.

The angular speed of the ball must decrease, and so the torque is provided by a friction force that acts up the hill.

b) The friction force results in an angular acceleration, related by I? = fR. The equation of motion is mg sin ? – f = ma and the acceleration and angular acceleration cm, are related by a = R? (note that positive acceleration is taken to be down the incline, cm and relation between a and ? is correct for a friction force directed uphill). cm Combining, ?I?() mg sin? = ma?1 + ? = ma 7 5 , ? mR2 ? from which a = (5 7)g sin?. c) From either of the above relations between if f and cm a, cm 22 f = ma = mg sin ? ? ?Á n = ?Á mg cos ?, cm s s 57 s

10.27: a) ? = ? t = (FR I ) t = ((18.0 N)(2.40 m) (2100 kg ? m2 ))(15.0 s) = 0.3086 rad/s, 2

ave P (175 hp)(746 W / hp) 10.28: a) ? = = = 519 N ? m. ? (2400 rev/min)?? ? rad/s ?? ? 30 rev/min ? b)W=??=(519N?m)(2?)=3261J.

? 10.29: a) ? = I? = I t (12)(1.50kg)(0.100m)2)(1200revmin)???rads?? ? 30 rev min ? = 2.5 s = 0.377 N ? m.

(600 rev/min)(2.5 s) ave 60 s/min 1 d) K = I?2 2

1 ((1/ 2)(1.5 kg)(0.100 )?? ? ? rad/s ??2 = m)2 (1200 rev/min)? ? ? 2 ? ? 30 rev/min ?? = 59.2 J, the same as in part (c).

10.30: From Eq. (10.26), the power output is ? 2? rad/s ? P = ?? = (4.30 N ? m)? 4800 rev/min ?ù ? = 2161 W, ? 60 rev/min ? which is 2.9 hp.

10.31: a) With no load, the only torque to be overcome is friction in the bearings (neglecting air friction), and the bearing radius is small compared to the blade radius, so any frictional torque could be neglected.

22 ? 1950 N ? m I 42.2 kg ? m2 c) From either W = K = 1 ? 2 or Eq. (10.24), 2 W=??=(1950N.m)(5.00rev?ù2?rad/rev)=6.13?ù104J.

d), e) The time may be found from the angular acceleration and the total angle, but the instantaneouspowerisalsofoundfromP=??=105kW(141hp).Theaveragepoweris half of this, or 52.6 kW.

? ? ? rad/s ? ? ? ? 30 rev/min ? ? b)IfthetensionintheropeisF,F=wandsow=?/R=1.79?ù103N.

c) Assuming ideal efficiency, the rate at which the weight gains potential energy is the power output of the motor, or wv = P, so v = P w = 83.8 m/s. Equivalently, v = ?R.

10.34: As a point, the woman’s moment of inertia with respect to the disk axis is mR 2 , and so the total angular momentum is ?1 ? L = L + L = (I + I )? = ? M + m?R2? disk woman disk woman ? 2 ? ?1 ? = ? 110 kg + 50.0 kg ?(4.00 m)2 (0.500 rev/s ?ù 2? rad/rev) ?2 ? =5.28?ù103kg?m2/s.

10.35: a) mvr sin? = 115 kg ? m2 / s, with a direction from the right hand rule of into the page.

b)dLdt=?=(2kg)(9.8Nkg)?(8m)? (90??-36.9??)= ? = ? 2 sin 125 N m 125 kg m2 s , out of the page.

a) L = (mr2)(v r) = mvr L=(5.97?ù1024kg)(2.98?ù104ms)(1.50?ù1011m)=2.67?ù1040kg?m2s b) L = (2 5mr 2 )(?) L=(25)(5.97?ù1024kg)(6.38?ù106m)2(2?rad(24.0hr?ù3600shr)) = 7.07 ?ù1033 kg ? m2 s

10.37: The period of a second hand is one minute, so the angular momentum is

M 2? L = I? = l2 3T ? 6.0 ?ù10-3 kg ? 2? = ? ? ?ù -2 2 = ?ù -6 ? 2 ? 3 ? 60 s

10.38: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be proportional to the inverse of the square of the radius, and the final angular velocity is

? ? 2 ? 2? ?? ? 2 R rad 7.0 ?ù105 km 21 ? R ? ? (30 d)(86,400 s d ?? 16 km ? 2

10.39: a) The net force is due to the tension in the rope, which always acts in the radial direction, so the angular momentum with respect to the hole is constant. 22 2 11222 122112 22 11 d) No other force does work, so 1.03?ù10-2 J of work were done in pulling the cord.

1 I=(0.400kg?m2)+(8.00kg)(1.80m)2=2.56kg?m2, 1 2 and her final moment of inertia is 2

Then from Eq. (10.33), I 2.56 kg ? m2 2 1 I 0.9 kg ? m2 2

10.41: If she had tucked, she would have made (2) (3.6 kg ? m2 ) 18 kg ? m2 ) = 0.40 rev in the last 1.0 s, so she would have made (0.40 rev)(1.5 1.0) = 0.60 rev in the total 1.5 s.

10.42: Let I =I =1200kg?m2, 10 20 Then, from Eq. (10.33),

I (1 2)MR2 1 ? =? 1 =? =? 2 1 I + mR2 1 (1 2)MR2 + mR2 1 + 1 2m M 0

3.0 rad s = = 1.385 rad s 1 + 2(70) 120 or 1.39 rad s to three figures b)K=(12)(12)(120kg)(2.00m)2(3.00rads)2=1.80kJ,and K = (1 2)(I 1 + (70 kg)(2.00 m)2 )?2 = 499 J. In changing the parachutist’s horizontal 20 2 component of velocity and slowing down the turntable, friction does negative work.

L mv(l 2) ?= = I (1 3)Ml 2 + m(l 2)2 (0.500 kg)(12.0 m s)(0.500 m) = = 0.223 rad s. (1 3)(40.0 kg)(1.00 )2 + (0.500 )(0.500 )2 m kg m

Ignoring the mass of the mud in the denominator of the above expression gives ? = 0.225 rad s , so the mass of the mud in the moment of inertia does affect the third significant figure.

r 10.45: Apply conservation of angular momentum L, with the axis at the nail. Let object 1

Just after the bug jumps, it has angular momentum in one direction of rotation and the B

(a) Conservation of angular momentum: m v d = -m vd + 1 m L2? 10 1 3 2

1 ? 90.0 N ? (3.00 kg)(10.0 m s) (1.50 m) = -(3.00 kg)(6.00 m s)(1.50 m) + ? ? (2.00 m)2 ? 3 ? 9.80 m s2 ? ? = 5.88 rad s

(b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved.

vertical force on the gyroscope, so the force that the pivot exerts must be equal in magnitude to the weight of the gyroscope, F = ? = mg = (0.165 kg)(9.80 m s2 ) = 1.617 N, 1.62 N to three figures. b) Solving Eq. (10.36) for ?, ?R (1.617N)(4.00?ù10-2m) ?= =( )( =188.7rads, I 1.20?ù10-4kg?m2 2?)rad2.20s whichis1.80?ù103revmin.Notethatinthisandsimilarsituations,since appearsin the denominator of the expression for ? , the conversion from rev s and back to rev min must be made.

c) K (1/2)((1/2)MR2)?2 10.49: a) = PP (1/ 2)((1/ 2)(60,000 kg)(2.00 m)2 )((500 rev/min)( ? rad/s ))2 = 30 rev/min 7.46 ?ù104 W = 2.21?ù103 s, or 36.8 min.

b) ? = I ? ? ? rad/s ? ? 2? rad ? = (1/ 2)(60,000 kg)(2.00 m)2 (500 rev/min)? ?(1.00??/s)? ? ? 30 rev/min ? ? 360?? ? = 1.10 ?ù105 N ? m.

10.50: Using Eq. (10.36) for all parts, a) halved b) doubled (assuming that the added weight is distributed in such a way that r and I are not changed) c) halved (assuming that w and r are not changed) d) doubled e) unchanged.

86,400 s = 2? and the mass and radius of the earth from Appendix F, (26,000 y)(3.175?ù107 s/y) ?~5.4?ù1022N?m.

10.52: a) The net torque must be ? 2? rad/s ? ?120 rev/min ?ù ? ? = I? = I ? = (1.86 kg ? m2 ) ? 60 rev/min ? = 2.60 N ? m. t (9.00 s) This torque must be the sum of the applied force FR and the opposing frictional torques ? at the axle and fr = ?Á nr due to the knife. Combining, fk

1 F = (? + ? + ?Á nr) fk R = 1 ((2.60 N ? m) + (6.50 N ? m) + (0.60)(160 N)(0.260 m)) 0.500 m = 68.1 N.

b) To maintain a constant angular velocity, the net torque ? is zero, and the force F ? is F? = 1 (6.50 N ? m + 24.96 N ? m) = 62.9 N. c) The time t needed to come to a stop is 0.500 m f

L ?I (120 rev/min ?ù 2? rad/s ) (1.86 kg ? m2 ) t = = = 60 rev/min = 3.6 s. ? ? (6.50 N ? m) ff

2? 2s 2(5.00 m) c) The work done by the rope on the flywheel will be the final kinetic energy; K=W=Fs=(40.0N)(5.0m)=200J.

2K 2(200 J) d) I = = =1.44kg?m2. ? 2 (16.67 )2 rad/s

??? ?t? ?I? ?I? 2 20.0 d) From the result of part (c), the power is (500 W)( 6.00 ) = 2.6 kW. e) No; the 3/2 20.00 power is proportional to the time t or proportional to the square root of the angle.

10.56: a) From the right-hand rule, the direction of the torque is i^ ?ù ^j = k^, the + z direction.

b), c) d) The magnitude of the torque is F (x – x2 l), which has it maximum at l 2. The 0

? (? I ) ? The angle in radiants is ? 2, the moment of inertia is

(1 3) ((750 N) (9.80 m s2 )(1.25 m))3 = 39.9 kg ? m2 and the torque is (220 N)(1.25 m) = 275 N ? m. Using these in the above expression gives t 2 = 0.455 s 2 , so t = 0.675 s.

10.58: a) From geometric consideration, the lever arm and the sine of the angle r between F and r are both maximum if the string is attached at the end of the rod. b) In terms of the distance x where the string is attached, the magnitude of the torque is Fxh x 2 + h 2 . This function attains its maximum at the boundary, where x = h, so the string should be attached at the right end of the rod. c) As a function of x, l and h, the torque has magnitude (x-l2)2+h2

This form shows that there are two aspects to increasing the torque; maximizing the lever arm l and maximizing sin ?. Differentiating ? with respect to x and setting equal to zero givesx =(l2)(1+(2hl)2).Thiswillbethepointatwhichtoattachthestringunless max 2h > l, in which case the string should be attached at the furthest point to the right, x = l.

b) In this case I = (4 3)ML2 and the gravitational torque is c) In this case I = (1 3)ML2 and the gravitational torque is d) The greater the angular acceleration of the upper end of the cue, the faster you would have to react to overcome deviations from the vertical.

?2 ?0 a) W = FRcos? d ? = FR

b) In Eq. (6.14), dl is the horizontal distance the point moves, and so W = F ? dl = FR, thesameaspart(a).c)FromK=W=(MR24)?2,?=4FMR.d)The 2

torque, and hence the angular acceleration, is greatest when ? = 0, at which point ? = (? I ) = 2F MR , and so the maximum tangential acceleration is 2F M . e) rad

10.61: The tension in the rope must be m(g + a) = 530 N. The angular acceleration of the cylinder is a R = 3.2 rad/s2 , and so the net torque on the cylinder must be 9.28 N ? m. Thus, the torque supplied by the crank is (530 N)(0.25 m) + (9.28 N ? m) = 141.8 N ? m, and the force applied to the crank handle is 0.12 m

10.62: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero, so balancing torques would not be correct. Balancing vertical forces, F cos? = f + w + F , and balacing horizontal forces rod F sin ? = n. With f = ?Á n, these equations become rod k

F cos? = ?Á n + F + w, rod k rod (a) Eliminating n and solving for F gives rod ? + F (16.0 kg) (9.80 m/s2 ) + (40.0 N) F = = = 266 N. cos? – ?Á sin? 30?? – 30?? rod cos (0.25) sin k

b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is (F – f )R, and f = ?Á n may be found k

k The acceleration of the block and the angular acceleration of the pulley are related by ? = ?R.

a) Multiplying the first of these relations by I R and eliminating ? in terms of a, and then adding to the second to eliminate T gives (sin ? – ?Á cos ?) g(sin ? – ?Á cos ?) a = mg k = (1 mkR2 ) , m + I / R2 + I /

and substitution of numerical values given 1.12 m/s 2 . b) Substitution of this result into either of the above expressions involving the tension gives T = 14.0 N.

10.64: For a tension T in the string, mg – T = ma and TR = I? = I a . Eliminating T and R

solving for a gives mg a=g = , m + I / R 2 1 + I / mR 2

where m is the mass of the hanging weight, I is the moment of inertia of the disk combination(I=2.25?ù10-3kg?m2fromProblem9.89)andRistheradiusofthediskto The acceleration is larger in case (b); with the string attached to the larger disk, the tension in the string is capable of applying a larger torque.

10.65: Taking the torque about the center of the roller, the net torque is fR = ?I , I = MR2 for a hollow cylinder, and with ? = a / R, f = Ma (note that this is a relation ?? between magnitudes; the vectors f and a are in opposite directions). The net force is F – f = Ma, from which F = 2Ma and so a = F 2M and f = F 2 .

cord does not slip, the angular acceleration of the pulley will be ? = a . Denoting the R

tensions in the cord as T and T , the equations of motion are AB

m g -T = m a AAA T-mg=ma BBB I T – T = a, AB2 R where the last equation is obtained by dividing ? = I? by R and substituting for ? in terms of a.

Adding the three equations eliminates both tensions, with the result that m -m a=g A B m + m + I / R2 AB Then, R mR+mR+I/R AB The tensions are then found from 2m m + m I R2 T = m (g – a) = g A B A AA ++2 m m IR AB 2m m + m I R2 BB ++2 m m IR AB

K = MgL sin? , from which 1 3v 2 3(2.50 m s) 2 4gsin? 4(9.80ms2)sin30.0??

This same result may be obtained by an extension of the result of Exercise 10.26; for the b) Both the translational and rotational kinetic energy depend on the mass which cancels the mass dependence of the gravitational potential energy. Also, the moment of inertia is proportional to the square of the radius, which cancels the inverse dependence of the angular speed on the radius.

10.68: The tension is related to the acceleration of the yo-yo by (2m)g – T = (2m)a, and to the angular acceleration by Tb = I? = I a . Dividing the second equation by b and b

adding to the first to eliminate T yields 2m 2 2 a=g =g , ?=g , (2m+Ib2) 2+(Rb)2 2b+R2b

where I = 2 1 mR 2 = mR 2 has been used for the moment of inertia of the yo-yo. The 2

tension is found by substitution into either of the two equations; e.g.,

radius of the path of the center of mass of the marble is R – r, so the condition that the ballstayonthetrackisv2=g(R-r).Thespeedisdeterminedfromthework-energy theorem,mgy=(12)mv2+(12)I?2.Atthispoint,itiscrucialtoknowthatevenforthe curved track, ? = v r; this may be seen by considering the time T to move around the circle of radius R – r at constant speed V is obtained from 2? (R – r) = Vt, during which timethemarblerotatesbyanangle2?(R-1)=?T,fromwhich?=Vr.Thework- r

energy theorem then states mgy = (7 10)mv 2 , and combining, canceling the factors of m and g leads to (7 10)(R – r) = h + r – 2R, and solving for h gives h=(2710)R-(1710)r.b)Intheabsenceoffriction,mgy=(12)mv2,andsubstitution of the expressions for y and v 2 in terms of the other parameters gives (1 2)(R – r) = h – r – 2R, which is solved for h = (5 2)R – (3 2)r.

? 10.70: In the first case, F and the friction force act in opposite directions, and the friction force causes a larger torque to tend to rotate the yo-yo to the right. The net force to the right is the difference F – f , so the net force is to the right while the net torque causes a clockwise rotation. For the second case, both the torque and the friction force tend to turn the yo-yo clockwise, and the yo-yo moves to the right. In the third case, friction tends to move the yo-yo to the right, and since the applied force is vertical, the yo-yo moves to the right.

T 1 1 ?a ? PR = MR2? = MR2? T ? 2 2 ?R? 2P 200 N a = = = 50 m/s2 T M 4.00 kg Distance the cable moves: x = 1 at 2 2

1 = 2 v=v+at=0+(50m/s2)(1.41s)=70.5ms 0 (b) For a hoop, I = MR 2 , which is twice as large as before, so ? and a would be T

half as large. Therefore the time would be longer. For the speed, v 2 = v 2 + 2ax, in which 0

x is the same, so v would be smaller since a is smaller 10.73: Find the speed v the marble needs at the edge of the pit to make it to the level ground on the other side. The marble must travel 36 m horizontally while falling 0y y 0

y – y = v t + 1 a t 2 gives t = 2.02 s 0 0y 2 y 0 0x 0x

Use conservation of energy, where point 1 is at the starting point and point 2 is at the edge of the pit, where v = 17.82 m/s. Take y = 0 at point 2, so y = 0 and y = h. 21 K +U = K +U 1122 mgh = 1 mv 2 + 1 I? 2 22 Rollingwithoutslippingmeans?=vr.I=2mr2,so1I?2=1mv2 525 mgh = 7 mv2 10 7v 2 7(17.82 m/s) h = = = 23 m 10g 10(9.80m/s2) 25 c) All is the same, except there is no rotational kinetic energy term in K : K = 1 mv 2 2

Rough : mgh = 1 mv + 1 I? 2 1222 1 1 ? 2 ?? v ? 2 mgh = mv2 + ? mR2 ?? ? 1 2 2 ? 5 ?? R ? 10 v 2 = gh 1 7 Smooth: Rotational KE does not change.

2 leaves the top of the cliff. Let point 1 be at the bottom of the hill and point 2 be at the top of the hill. Take y = 0 at the bottom of the hill, so y = 0 and y = 28.0 m. 12 K +U = K +U 1122 1 mv 2 + 1 I? 2 = mgy + 1 mv 2 + 1 I? 2 2121 22222 Rollingwithoutslippingmeans?=vrand1I?2=1(2mr2)(v/r)2=1mv2 2 25 5 7 mv 2 = mgy + 7 mv 2 10 1 2 10 2

v = v 2 – 10 gy = 15.26 m s 2172 Consider the projectile motion of the ball, from just after it leaves the top of the cliff Use the vertical motion to find the time in the air: 0y y 0

y – y = v t + 1 a t 2 gives t = 2.39 s 0 0y 2 y 0 0x

Just before it lands, v = v + a t = 23.4 s and v = v = 15.3 s v = v2 + v2 = 28.0 m s y 0y y x 0x x y b) At the bottom of the hill, ? = v r = (25.0 m s)r. The rotation rate doesn’t change while the ball is in the air, after it leaves the top of the cliff, so just before it lands ? = (15.3 s)r. The total kinetic energy is the same at the bottom of the hill and just before it lands, but just before it lands less of this energy is rotational kinetic energy, so the translational kinetic energy is greater.

22 ?1 ? I = I + I = M R2 + 6? m R2 ? rim spokes r ? 3 s ? Uniform density means: m = ? 2?R and m = ?R. No slipping means that ? = v R. rs Also, m = m + m = 2?R? + 6R? = 2R?(? + 3) substituting into (1) gives rs

2R?(?+3)gh=1(2R?)(?+3)(R?)2+1?2?R?R2+ ?1?RR2???2 ? 6? ?? 2 2 ? ? 3 ?? (? + 3)gh (? + 3)(9.80 m s2 )(58.0 m) ? = = (0.210 m)2 (? + 2) = 124 rad s R2 (? + 2) and v = R? = 26.0 m s

(b) Doubling the density would have no effect because it does not appear in the answer. ? ? 1 , so doubling the diameter would double the radius which would reduce R

? = r? = (0.330 m)(2? rad s) = 2.07 s b) ? = v r = (2.07 m s) (0.655 m = 3.16 rad s = 0.503 rev s c) ? = v r = (2.07 m s) (0.220 m) = 9.41 rad s = 1.50 rev s

10.78: a) The kinetic energy of the ball when it leaves the tract (when it is still rolling without slipping) is (7 10)mv2 and this must be the work done by gravity, W = mgh, so v = 10gh 7. The ball is in the air for a time t = 2y g, so x = vt = 20hy 7. b) The answer does not depend on g, so the result should be the same on the moon. d)Forthedollarcoin,modeledasauniformdisc,K=(34)mv2,andsox=8hy3.

(10)(0.800) (1 2) (400 ) (0.15 )2 10K N m m 10.79: a) v = = = 9.34 m s. 7m 7(0.0590 kg)

b) Twice the speed found in part (a), 18.7 m s. c) If the ball is rolling without slipping, the speed of a point at the bottom of the ball is zero. d) Rather than use the intermediate calculation of the speed, the fraction of the initial energy that was converted to gravitational potential energy is (0.800) (0.900), so (0.720) (1 2)kx2 = mgh and solving for h gives 5.60 m.

10.80: a) b) R is the radius of the wheel (y varies from 0 to 2R) and T is the period of the c) Differentiating,

2?R ? ? 2?t ?? ? 2? ?2 ? 2?t ? v = ?1 – cos? ?? a = ? ? R sin? ? x ? ? ?? x ? ? ? ? TTTT 2?R ? 2?t ? ? 2? ?2 ? 2?t ? y??y???? TTTT

? 2?t ? d) vx = vy = 0 when ? ? = 2? or any multiple of 2?, so the times are integer ?T? multiples of the period T. The acceleration components at these times are 4? 2 R x y T2 ? 2? ? 2 ? 2?t ? ? 2?t ? 4? 2 R e) a 2 + a 2 = ? ? R cos 2 ? ? + sin 2 ? ? = , x y ? T ? ? T ? ? T ? T2

independent of time. This is the magnitude of the radial acceleration for a point moving on a circle of radius R with constant angular velocity 2? . For motion that consists of this T

circular motion superimposed on motion with constant velocity ( r the acceleration a = 0), due to the circular motion will be the total acceleration.

initially, this is 32.0 J and at the return to the bottom it is 8.0 J. Friction has done – 24.0 J of work, – 12.0 J each going up and down. The potential energy at the highest point was 20.0 J (0.600 kg ) 9.80 m s 2

10.82: Differentiating , and obtaining the answer to part (b),

d? ? ? ?2 3 ?= =3bt2=3b? ? =3b13?23, dt ? b ? d? ? ? ?1 3 dt ? b ?

a) W=?I ?d?=6b23I ??13d?=9I b23?43. cm cm cm 2 c) The kinetic energy is K = 1 I ? 2 = I b 2 3? 4 3 , 9 cm cm 22 in agreement with Eq. (10.25); the total work done is the change in kinetic energy.

10.83: Doing this problem using kinematics involves four unknowns (six, counting the two angular accelerations), while using energy considerations simplifies the calculations greatly. If the block and the cylinder both have speed v, the pulley has angular velocity v/R and the cylinder has angular velocity v/2R, the total kinetic energy is

1? M(2R)2 MR2 ? 3 2? 2 2 ? 2 This kinetic energy must be the work done by gravity; if the hanging mass descends a distance y, K = Mgy, or v2 = (2 3)gy. For constant acceleration, v2 = 2ay, and comparison of the two expressions gives a = g 3.

?? = I? 11 mg cos? = mL2? 23 3 cos? ? 3 ? (9.80 m s2 ) cos 60?? ? = g = ? ? = 0.92 rad s2 2 L ? 2 ? 8.00 m (b) As the bridge lowers, ? changes, so ? is not constant. Therefore Eq. (9.17) is not valid.

(c) Conservation of energy: 1 PE = KE ? mgh = I?2 if 2 1?1 ? mg L sin ? = ? mL2 ??2 2 2?3 ? 3g sin? ?= L 3(9.8m s2)sin60?? = = 1.78 rad s 8.00 m

10.85: The speed of the ball just before it hits the bar is v = 2gy = 15.34 m s. Use conservation of angular momentum to find the angular velocity ? of the bar just L=mvr=(5.00kg)(15.34ms)(2.00m)=153.4kg?m2 1

Immediately after the collsion the bar and both balls are rotating together. L =I ? 2 tot

1 1 ( )( ) ( )( ) I = Ml2+2mr2= 8.00kg 4.00m +25.00kg 2.00m =50.67kg?m2 22 tot 12 12 L = L = 153.4 kg ? m2 21 ? = L I = 3.027 rad s 2 tot Just after the collision the second ball has linear speed v=rw=(2.00m)(3.027rads)=6.055msandismovingupward.

torque on the system, and so the angular momentum will be constant. As the rings slide toward the ends, the moment of inertia changes, and the final angular velocity is given by Eq. (10.33), ? 1 ML2 + 2mr 2 ? ?ù -4 ? 2 I 5.00 10 kg m ? ? = ? 1 = ? ? 12 1 ? = ? = 1 , I ? 1 ML + 2mr ? 2.00 ?ù10- kg ? m 4 21122132 2 12 2 2

b) The forces and torques that the rings and the rod exert on each other will vanish, but the common angular velocity will be the same, 7.5 rev/min.

10.87:Theintialangularmomentumofthebulletis(m4)(v)(L2),andthefinalmoment of intertia of the rod and bullet is (m 3)L2 + (m 4)(L 2) = (19 48)mL2. Setting the initial 2

mvL 8 6 angular moment equal to ?I and solving for? gives ? = = v L. (19 48)mL2 19 (1 2)I?2 (19 )mL2 ((6 19) (v ))2 48 L 3 b) = = . (1 2)(m 4)v2 (m 4)v2 19

10.88: Assuming the blow to be concentrated at a point (or using a suitably chosen ÔÇ£averageÔÇØ point) at a distance r from the hinge, ?? = rF and L = rF t = rJ. ave ave, ave The angular velocity ? is then L rF t (l 2)F t 3 F t ? = = ave = ave = ave , I I 1 ml 2 2 ml 3

Where l is the width of the door. Substitution of the given numeral values gives ? = 0.514 rad s.

10.89: a) The initial angular momentum is L = mv(l 2) and the final moment of inertia is I = I + m(l 2)2 , so 0

00 22 common mass the moment of inertia is proportional to the square of the radius, R2?=R2?,orR2?=(R+R)(?+?)~R2?+2RR?+R2?, 2 0022000000000 where the terms in R ? and ? 2 have been omitted. Canceling the R 2? term gives 00 R? 2? 0

10.91: The initial angular momentum is L = ? I and the initial kinetic energy is 1 0A K = I ?2 2. The final total moment of inertia is 4I , so the final angular velocity 1 A0 A is (1 4)? and the final kinetic energy is ( ) (? 4) = (1 4)K . (This result may be 2 1 2 4I 0 A0 1 obtainedmoredirectlyfromK=L2I.Thus,K=-(34)KandK=-(43)(-2400J) 11 = 3200 J.

10.92: The tension is related to the block’s mass and speed, and the radius of the circle, v2 byT = m . The block’s angular momentum with respect to the hole is L = mvr , so in r terms of the angular momentum,

m2v2 2 (mvr)2 L2 1r r m r 3 mr 3 mr 3

The radius at which the string breaks can be related to the initial angular momentum by L2 (mvr) ((0.250kg)(4.00ms)(0.800m)) 22 r3 = = 1 1 = , mT mT (0.250kg)(30.0N) max max from which r = 0.440 m.

10.93: The train’s speed relative to the earth is 0.600 m s + ? (0.475 m) , so the total angular momentum is ((0.600 ) ?(0.475 )) (1.20 )(0.475 ) (1 2) (7.00 ) ? 1.00 m ?2 m s + m kg m + ? kg ? ? = 0, ?2?

from which ? = -0.298 rad s ,with the minus sign indicating that the turntable moves clockwise, as expected.

rrrr b) Using the vector product form for the angular momentum, v = -v and r = -r , so 1212

rrrr mr ?ù v = mr ?ù v , 2211 r so the angular momenta are the same. c) Let ? = ?^. Then, j rrr() v = ? ?ù r = ? zi^ – xk^ , and L=mr?ùv=m?-xR)i^+(x2+y2)^j+(xR 111

r rr Withx2+y2=R2,themagnitudeofLis2m?R2,andL??=m?2R2,andso 1 r cos? = m? R = 1 , and ? = ? . This is true for L as well, so the total angular 22

(2m?R2)(?) 2 6 2 momentum makes an angle of ? with the +y-axis. d) From the intermediate 6

calculation of part (c), L = m?R2 = mvR, so the total y-component of angular y1 momentum is L = 2mvR. e) L is constant, so the net y-component of torque is zero. f) yy Each particle moves in a circle of radius R with speed v, and so is subject to an inward force of magnitude mv2 R. The lever arm of this force is R, so the torque on each has magnitude mv2. These forces are directed in opposite directions for the two particles, and the position vectors are opposite each other, so the torques have the same magnitude and direction, and the net torque has magnitude 2mv2 .

final total moment of inertia is I + mr2, so the final angular velocity is b) The kinetic energy after the collision is 1( )( ) K = ?2 mr2 + I = M + m gh, or 2 2(M + m)gh ? = (mr 2 + I ) .

? m ?( r ), c) Substitution of ? = ?r 2 into either of the result of part (a) gives ? = ? ? v ?m+M? andintotheresultofpart(b),?= 2gh(1r),whichareconsistentwiththeformsforv.

10.96: The initial angular momentum is ?? – mRv , with the minus sign indicating that 11 runner’s motion is opposite the motion of the part of the turntable under his feet. The final angular momentum is ? (? + mR 2 ), so 2

?? – mRv ?= 1 1 2 ? + mR2 (80 kg ? m2 )(0.200 rad s) – (55.0 kg)(3.00 m)(2.8 m s) = (80 kg ? m2 ) + (55.0 kg)(3.00 m)2 = -0.776 rad s, where the minus sign indicates that the turntable has reversed its direction of motion (i.e., the man had the larger magnitude of angular momentum initially).

10.97: From Eq. (10.36), ?r (50.0 kg)(9.80 m s2 )(0.040 m) = = = 12.7 rad s, ?? (0.085 kg ? m2 )((6.0 m s) (0.33 m))

cm m velocity will change by ? = – cm ? The change is velocity of the end of the bat will J(x x ) I

J J(x-x )x then be v = v _ ?x = – cm cm ? Setting end cm cm m? v=0allowscancellationofJ,andgives?=(x-x)xm,whichwhensolvedforx end cm cm is ? (5.30?ù10-2kg?m2) cm x m (0.600 m)(0.800 kg) cm

10.99: In Fig. (10.34(a)), if the vector r, and hence the vector L are not horizontal but make an angle ? with the horizontal, the torque will still be horizontal (the torque must be perpendicular to the vertical weight). The magnitude of the torque will be ? r cos ? , and this torque will change the direction of the horizontal component of the angular momentum, which has magnitude L cos ? .

Thus, the situation of Fig. (10.36) is reproduced, but with Lhoriz instead of L . Then, the expression found in Eq. (10.36) becomes

The distance from the center of the ball to the midpoint of the line joining the points where the ball is in contact with the rails is R2 – (d 2) , so v = ? R2 – d 2 2 cm when d = 0, this reduces to v = ?R, the same as rolling on a flat surface. When cm cm

11 b) K = mv2 + I?2 22 ? ? ?2 ? = 1 ?mv2 + (2 5)mR2? vcm ? ? 2 ? ? R2 – (d 2 4) ? ? cm c) ? ? ? ? mv2 ? 2 ? 10 ? 1 – d 2 4R2 ? Setting this equal to mgh and solving for v gives the desired result. c) The cm denominator in the square root in the expression for v is larger than for the case cm d = 0, so v is smaller. For a given speed, ? is large than the d = 0 case, so a larger cm fraction of the kinetic energy is rotational, and the translational kinetic energy, and hence v , is smaller. d) Setting the expression in part (b) equal to 0.95 of that of the d = 0 cm case and solving for the ratio d R gives d R = 1.05. Setting the ratio equal to 0.995 gives d R = 0.37.

kk k acceleration is fR = k = v = at = ? ?Á MgR 2 ?Á g k . b) Setting R I (1 2)MR2 R gives R? R? R? t= 0 = 0 = 0, a + R? ?Á g + 2?Á g 3?Á g kkk and

d = 1 at 2 = 1 (?Á g )?? R?0 ??2 = ?? R2?2 0 The magnitude of the angular = (? – ?t )R and solving for t 0

LR F and F must satisfy are LR F +F =w LR I? F -F = , LR r where the second equation is ? = L, divided by r. These two equations can be solved for the forces by first adding and then subtracting, yielding 1 ? I? ? F = ?? + ? L 2? r ? 1 ? I? ? R 2? r ? Usingthevalues?=mg=(8.00kg)(9.80ms2)=78.4Nand I? (8.00 kg)(0.325 m)2 (5.00 rev s ?ù 2? rad rev) = = 132.7 kg ? m s r (0.200 m) gives LR LR LR c) = 0.3 rev s = 1.89 rad s, F = 165 N, F = -86.2 N, LR R 66.4 N?s

rr 10.103:a)SeeProblem10.92;T=mv2r2r3.b)Tanddrarealwaysantiparallel,so 11

?r21 dr mv2 ? 1 1 ? ?r1 2 1 ? r2 r12 ? 1 1 r3 2

c) v = v (r r ), so 2112 ?? ?2 ? 1 mv2 r K=m(v2-v2)=1??1?-1?’ 2 1 ??? ? ?? 2 2r 2

(1.00 kg)(0) + (2.00 kg)(0.580 m) x = = 0.387 m cm 3.00 kg from the center of the small ball.

11.2: The calculation of Exercise 11.1 becomes (1.00 kg)(0) + (1.50 kg)(0.280 m) + (2.00 kg)(0.580 m) x = = 0.351 m cm 4.50 kg This result is smaller than the one obtained in Exercise 11.1.

11.3: In the notation of Example 11.1, take the origin to be the point S, and let the child’s distance from this point be x. Then, M (- D 2) + mx MD s = = 0, x = = 1.125 m, cm M + m 2 m which is (L 2 – D 2) 2, halfway between the point S and the end of the plank.

11.4: a) The force is applied at the center of mass, so the applied force must have the same magnitude as the weight of the door, or 300 N. In this case, the hinge exerts no b) With respect to the hinge, the moment arm of the applied force is twice the distance to the center of mass, so the force has half the magnitude of the weight, or 150 N . The hinge supplies an upward force of 300 N – 150 N = 150 N.

11.6: The other person lifts with a force of 160 N – 60 N = 100 N. Taking torques about the point where the 60 – N force is applied, ? 160 N ? ? 100 N ?

11.7: If the board is taken to be massless, the weight of the motor is the sum of the applied forces, 1000 N. The motor is a distance = 1.200 m from the end where (2.00 m)(600 N) (1000 N) the 400-N force is applied.

to do this problem, a sneaky way is to say that the lifters each exert 100 N to the lift the board, leaving 500 N and 300 N to the lift the motor. Then, the distance of the motor from the end where the 600-N force is applied is = 0.75 m .The center of (2.00 m)(300 N) (800 N)

gravity is located at + = 0.80 m from the end where the 600 N force (200 N)(1.0 m) (800 N)(0.75 m) (1000 N) is applied.

11.9: The torque due to T is – T h = – Lw cot ? h, and the torque due to T is T D = Lw . xxD yy ThesumofthesetorquesisLw(1-hcot?).FromFigure(11.9(b)),h=Dtan?,sothenet D

11.10: a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force. For the vertical forces to balance, n = w + w = 160 N + 740 N = 900 N, and the maximum frictional forces is 21m ?Á n = (0.40)(900 N) = 360 N (see Figure 11.7(b)). b) Note that the ladder makes contact s2 with the wall at a height of 4.0 m above the ground. Balancing torques about the point of contact with the ground, (4.0 m)n = (1.5 m)(160 N) + (1.0 m)(3 5))(740 N) = 684 N ? m, 1

so n = 171.0 N, keeping extra figures. This horizontal force about must be balanced by 1 the frictional force, which must then be 170 N to two figures. c) Setting the frictional force, and hence n , equal to the maximum of 360 N and solving for the distance x along 1

the ladder, (4.0 m)(360 N) = (1.50 m)(160 N) + x(3 5)(740 N), so x = 2.70 m, or 2.7 m to two figures.

11.11: Take torques about the left end of the board in Figure (11.21). a) The force F at the support point is found from F(1.00m)=+(280N)(1.50m)+(500N)(3.00m),orF=1920N.b)Thenetforcemustbe zero, so the force at the left end is (1920 N) – (500 N) – (280 N) = 1140 N, downward.

b) x = 6.25 m when F = 0, which is 1.25 m beyond point B. c) Take torques about A

the right end. When the beam is just balanced, F = 0, so F = 900 N. The distance that AB (300 N)(4.50 m) (900 N)

11.13: In both cases, the tension in the vertical cable is the weight ?. a) Denote the length of the horizontal part of the cable by L. Taking torques about the pivot point, TLtan30.0??=wL+w(L2),fromwhichT=2.60w.Thepivotexertsanupwardvertical force of 2w and a horizontal force of 2.60w , so the magnitude of this force is 3.28w , directed 37.6?? from the horizontal. b) Denote the length of the strut by L , and note that the angle between the diagonal part of the cable and the strut is 15.0??. Taking torques about the pivot point, TL sin 15.0?? = wL sin 45.0?? + (w 2)L sin 45??, so T = 4.10w. The horizontal force exerted by the pivot on the strut is then T cos 30.0?? = 3.55? and the vertical force is (2w) + T sin 30?? = 4.05w, for a magnitude of 5.38w, directed 48.8??.

11.14: a) Taking torques about the pivot, and using the 3-4-5 geometry,

other hinge; then, the vertical forces that the hinges exert have no torque. The horizontal force is found from F (1.00 m) = (280 N)(0.50 m), from which F = 140 N. The top hinge HH exerts a force away from the door, and the bottom hinge exerts a force toward the door. Note that the magnitudes of the forces must be the same, since they are the only horizontal forces.

11.16: (a) Free body diagram of wheelbarrow: ?? = 0 wheel – (450 N)(2.0 m) + (80 N)(0.70 m) + W (0.70 m) = 0 L

The tension in the guy wire is found from TL sin 60?? = (5000 N) L cos 60.0?? + (2600 N)(0.35 L) cos 60.0??,

so T = 3.14 kN. The vertical force exerted on the boom by the pivot is the sum of the ?F ? weights, 7.06 kN and the horizontal force is the tension, 3.14 kN. b) No; tan ? v ? ? 0. ? FH ?

11.19: To find the tension T in the left rope, take torques about the point where the L

rope at the right is connected to the bar. Then, T (3.00 m) sin 150?? = (240 N)(1.50 m) + (90 N)(0.50 m), so T = 270 N. The vertical LL component of the force that the rope at the end exerts must be (330 N) – (270 N) sin 150?? = 195 N, and the horizontal component of the force is – (270 N) cos150??, so the tension is the rope at the right is T = 304 N. and ? = 39.9??. R

11.20: The cable is given as perpendicular to the beam, so the tension is found by T (3.00 m) = (1.00 kN)(2.00 m) cos 25.0?? + (5.00 kN)(4.50 m) cos 25.0??, or T = 7.40 kN. The vertical component of the force exerted on the beam by the pivot is the net weight minus the upward component of T , 6.00 kN – T cos 25.0?? = 0.17 kN. The horizontal force is T sin 25.0?? = 3.13 kN.

11.21: a) F (3.00 m) – F (3.00 m + l) = (8.00 N)(-l). This is given to have a 12 magnitude of 6.40 N.m, so l = 0.80m. b) The net torque is clockwise, either by considering the figure or noting the torque found in part (a) was negative. c) About the pointofcontactofF2,thetorqueduetoF1is-Fl,andsettingthemagnitudeofthis 1

11.22: From Eq. (11.10), l (0.200 m) – Y = F 0 = F = F (1333 m 2 ). l? (3.0?ù10-2 m)(50.0?ù10-4 m2) Then,F=25.0NcorrespondstoaYoung’smodulusof3.3?ù104Pa,andF=500N corresponds to a Young’s modulus of 6.7 ?ù105 Pa.

Fl (400 N)(2.00 m) 11.23: A= 0= =1.60?ù10-6m2, Y l (20?ù1010 Pa)(0.25?ù10-2 m) andsod=4A?=1.43?ù10-3m,or1.4mmtotwofigures.

11.24: a) The strain, from Eq. (11.12), is l = F . For steel, using Y from Table (11.1) l YA 0

andA=?d2=1.77?ù10-4m2, 4 l l (2.0?ù1011 Pa)(1.77?ù10-4 m2) 0

Similarly,thestrainforcopper(Y=1.10?ù1011Pa)is2.1?ù10-4.b)Steel: (1.1?ù10-4)?ù(0.750m)=8.3?ù10-5m.Copper:(2.1?ù10-4)(0.750m)=1.6?ù10-4m?

11.25: From Eq. (11.10), (5000 N)(4.00 m) ?ù -4 2 ?ù -2 (0.50 10 m )(0.20 10 m)

11.26: From Eq. (11.10), (65.0kg)(9.80m s2)(45.0m) (?(3.5?ù10-3m)2)(1.10m)

11.27: a) The top wire is subject to a tension of (16.0 kg)(9.80 m s 2 ) = 157 N and henceatensilestrainof =3.14?ù10-3,or3.1?ù103 (157 N) – to two figures. The (20?ù1010Pa)(2.5?ù10-7m2) bottom wire is subject to a tension of 98.0 N, and a tensile strain of 1.96 ?ù10-3 , or ?ù -3 ?ù -3 = 2.0 10 to two figures. b) (3.14 10 )(0.500 m) 1.57 mm, (1.96?ù10-3)(0.500m)=0.98mm.

a) 2 1.6 106 Pa. b) 1.6 Pa 0.8 11.28: s ?(12.5?ù10-2 m)2 2.0?ù1010 Pa (2.50 m)

as great. c) The volume is inversely proportional to the bulk modulus for a given pressure change, so the volume change of the lead ingot would be four times that of the gold.

11.31: a) 250N =3.33?ù106Pa. b) (3.33?ù106Pa)(2)(200?ù10-4m2)=133kN. 0.75?ù10-4 m2

11.32: a) Solving Eq. (11.14) for the volume change, V = -kV P = – ?ù -11 -1 3 ?ù 8 -1.0?ù105 Pa) (45.8 10 Pa )(1.00 m )(1.16 10 Pa = -0.0531 m3.

b) The mass of this amount of water not changed, but its volume has decreased to 0.947 m3

(600cm3)(3.6?ù106Pa) 1 -10 -1 11.33: B = = 4.8?ù109Pa, k = = 2.1?ù10 Pa . (0.45cm3) B

11.34: a) Using Equation (11.17), || 2.4 AS [(.10 m)(.005 m)][7.5 ?ù1010 Pa]

11.35: The area A in Eq. (11.17) has increased by a factor of 9, so the shear strain for the larger object would be 1 9 that of the smaller.

11.37: A ?(0.92?ù10-3 m)2 (1.6?ù103)(20?ù1010Pa)(5?ù10-6m2)=1.60?ù10 -3 11.38: a) N. b) If this were the case, the wire would stretch 6.4 mm.

F (2.40?ù108 Pa)(3.00?ù10-4 m2) 3 11.39: a = tot = – 9.80 m s2 = 10.2 m s2. m (1200 kg)

A = 350 N = ?ù -7 2 d = A ? = 11.40: 7.45 10 m , so 4 0.97 mm. 4.7?ù108 Pa

11.41: a) Take torques about the rear wheel, so that f?d = ?x , or x = fd . cm cm b) (0.53)(2.46 m) = 1.30 m to three figures.

11.42: If Lancelot were at the end of the bridge, the tension in the cable would be (from taking torques about the hinge of the bridge) obtained from T (12.0 N) = (600 kg)(9.80 m s2 )(12.0 m) + (200 kg)(9.80 m s2 )(6.0 m), so T = 6860 N. This exceeds the maximum tension that the cable can have, so Lancelot is going into the drink. To find the distance x Lancelot can ride, replace the 12.0 m multiplying Lancelot’s weight by x and the tension T by T = 5.80 ?ù103 N and solve max for x;

Taking the clockwise direction as positive, and taking torques about the center of mass, Forces: – F – W + F = 0 tail wing

Torques: – (3.66 m)F + (.3 m)F = 0 tail wing

A shortcut method is to write a second torque equation for torques about the tail, and solve for the F : -(3.66 m)(6700 N) + (3.36 m)F = 0. This gives wing wing wing tail

Note that the rear stabilizer provides a downward force, does not hold up the tail of the aircraft, but serves to counter the torque produced by the wing. Thus balance, along with weight, is a crucial factor in airplane loading.

11.44: The simplest way to do this is to consider the changes in the forces due to the extra weight of the box. Taking torques about the rear axle, the force on the front wheels is decreased by 3600 N = 1200 N, so the net force on the front wheels 1.00 m 3.00 m is10,780N-1200N=9.58?ù103Ntothreefigures.Theweightaddedtotherearwheels is then 3600 N + 1200 N = 4800 N, so the net force on the rear wheels is b) Now we want a shift of 10,780 N away from the front axle. Therefore, 1.00 m 3.00 m

11.45: Take torques about the pivot point, which is 2.20 m from Karen and 1.65 m from Elwood. Then w (1.65 m) = (420 N)(2.20 m) + (240 N)(0.20 m), so Elwood Elwood weighs 589 N. b) Equilibrium is neutral.

123 The center of gravity is a distance x to the right of point O where cm

w (5.0 cm) + w (9.5 cm) + w (10.0 cm – l 2) x=1 2 3 cm w + w + w 123

(10.0cm)(5.0cm)+(8.0cm)(9.5cm)+l(10.0cm-l 2) (10.0 cm) + (8.0 cm) + l

Setting x = 0 gives a quadratic in l, which has as its positive root l = 28.8 cm. cm b) Changing the material from steel to copper would have no effect on the length l since the weight of each piece would change by the same amount.

rrv r 11.47: Let r ? = r – R ,where R is the vector from the point O to the point P. ii rrr The torque for each force with respect to point P is then ?? = r ??ù F , and so the net torque iii is

r (r r ) r ? ? =? r – R ?ù F iii rr rr =?r?ùF-?R?ùF ii i rrrr ii i In the last expression, the first term is the sum of the torques about point O, and the second term is given to be zero, so the net torques are the same.

r 11.48: From the figure (and from common sense), the force F is directed along the 1

length of the nail, and so has a moment arm of (0.0800 m) sin 60?? . The moment arm of r F is 0.300 m, so 2

r balance the applied force F , and so has magnitude 120.0 N and is to the left. Taking torques about point A, (120.0 N)(4.00 m) + F (3.00 m), so the vertical component is V

– 160 N , with the minus sign indicating a downward component, exerting a torque in a direction opposite that of the horizontal component. The force exerted by the bar on the hinge is equal in magnitude and opposite in direction to the force exerted by the hinge on the bar.

11.50: a) The tension in the string is w = 50 N, and the horizontal force on the bar 2

must balance the horizontal component of the force that the string exerts on the bar, and is equal to (50 N) sin 37?? = 30 N, to the left in the figure. The vertical force must be ? 50 N ? (50 N) cos 37?? + 10 N = 50 N, up. b) arctan ? ? = 59??. c) (30 N)2 + (50 N)2 = 58 N. ? 30 N ? d) Taking torques about (and measuring the distance from) the left end, (50 N)x = (40 N)(5.0 m) , so x = 4.0 m , where only the vertical components of the forces exert torques.

11.51: a) Take torques about her hind feet. Her fore feet are 0.72 m from her hind feet, and so her fore feet together exert a force of = 73.9 N, so each foot exerts a (190 N) (0.28 m) (0.72 m) force of 36.9 N, keeping an extra figure. Each hind foot then exerts a force of 58.1 N. b) Again taking torques about the hind feet, the force exerted by the fore feet is + = 105.1 N, so each fore foot exerts a force of 52.6 N and each hind (190N)(0.28m)(25N)(0.09m) 0.72 m foot exerts a force of 54.9 N.

11.52: a) Finding torques about the hinge, and using L as the length of the bridge and w and w for the weights of the truck and the raised section of the bridge, TB

TLsin70??=w(3L)cos30??+w(1L)cos30??,so T4 B2

(3m+1m)(9.80ms2)cos30?? sin 70?? b) Horizontal: T cos(70?? – 30??) = 1.97 ?ù105 N. Vertical: w + w – T sin 40?? TB = 2.46 ?ù105 N.

r 11.53: a) Take the torque exerted by F to be positive; the net torque is then 2

12 b) ? = -(14.0 N)(3.0 m)sin 37?? = -25.3 N ? m, keeping an extra figure, and 1

?=(14.0N)(4.5m)sin37??=37.9N?m,andthenettorqueis12.6N?m.Aboutpoint 2

P, ? = (14.0 N)(3.0 m)(sin 37??) = 25.3 N ? m, and 1 ? = (-14.0 N)(1.5 m)(sin 37??) = -12.6 N ? m, and the net torque is 12.6 N ? m. The 2

11.54: a) Take torques about the pivot. The force that the ground exerts on the ladder is giventobevertical,andF(6.0m)sin?=(250N)(4.0m)sin? V

+(750N)(1.50m)sin?,soF=354N.b)Therearenootherhorizontalforcesonthe V

ladder, so the horizontal pivot force is zero. The vertical force that the pivot exerts on the ladder must be (750 N) + (250 N) – (354 N) = 646 N, up, so the ladder exerts a downward force of 646 N on the pivot. c) The results in parts (a) and (b) are independent of ?.

11.55: a) V = mg + w and H = T . To find the tension, take torques about the pivot point. Then, denoting the length of the strut by L, ?2 ? ?2 ? ?L? T? L?sin?=w? L?cos?+mg? ?cos?,or ?3 ? ?3 ? ?6? ? mg ? ? 4?

b) Solving the above for w , and using the maximum tension for T , mg w = T tan ? – = (700 N) tan 55.0?? – (5.0 kg) (9.80 m s2 ) = 951 N. 4

Lower rod: ?? = 0 : (6.0 N)(4.0 cm) = A(8.0 cm) p A = 3.0 N ?F = 0 : T = 6.0 N + A = 6.0 N + 3.0 N = 9.0 N 3

Middle rod: ?? = 0 : B(3.0 cm) = (9.0 N)(5.0 cm) p B = 15 N ?F = 0 : T = B + T = 15 N + 9.0 N = 24 N 23

?? = 0, axis at hinge T(6.0m)(sin40??)-w(3.75m)(cos30??)=0 T = 760 N 11.58: (a)

?? = 0 Hinge T (3.5 m)sin 37?? = (45,000 N)(7.0 m) cos 37?? T = 120,000 N

a) ?? = 0, axis at lower end of beam ?L? T(sin20??)L=-mg? ?cos40??=0 ?2? 1 mg cos 40?? T = 2 = 2700 N sin 20?? b) Take +y upward.

? F = 0 gives n – w + T sin 60?? = 0 so n = 73.6 N y ? F = 0 gives f = T cos 60?? = 1372 N xs f 1372 N f = ?Á n, ?Á = s = = 19 sss n 73.6 N The floor must be very rough for the beam not to slip.

11.60: a) The center of mass of the beam is 1.0 m from the suspension point. Taking torques about the suspension point, w(4.00 m) + (140.0 N)(1.00 m) = (100 N)(2.00 m) (note that the common factor of sin 30?? has been factored out), from which w = 15.0 N.

b) In this case, a common factor of sin 45?? would be factored out, and the result would be the same.

(200 N)(2.50 m) + (600 N) ?ù (5.00 m) – T (5.00 m) = 0 . Therefore the y-component of y

the tension is T = 700 N . The x-component of the tension is then y

T = (1000 N)2 – (700 N)2 = 714 N . The height above the pole that the wire must be x

attached is (5.00 m) 700 = 4.90 m . b) The y-component of the tension remains 700 N and 714 the x-component becomes (714 N) = 795 N , leading to a total tension of 4.90 m 4.40 m

11.62: A and B are straightforward, the tensions being the weights suspended; AB To find T and T , a trick making use of the right angle where the strings join is available; CD use a coordinate system with axes parallel to the strings. Then, T = T cos 36.9?? = 0.470 N, T = T cos 53.1?? = 0.353 N, To find T , take torques about CB DB E T(1.000m)=Tsin36.9??(0.800m)+Tsin53.1??(0.200m) EDC + (0.120 kg)(9.80 m s2 )(0.500 m) = 0.833 N ? m, so T = 0.833 N.T may be found similarly, or from the fact that T + T must be the E F EF

total weight of the ornament. (0.180 kg)(9.80 m s2 ) = 1.76 N, from which T = 0.931 N. F

11.63: a) The force will be vertical, and must support the weight of the sign, and is 300 N. Similarly, the torque must be that which balances the torque due to the sign’s weight about the pivot, (300 N)(0.75 m) = 225 N ? m . b) The torque due to the wire must balance the torque due to the weight, again taking torques about the pivot. The minimum tension occurs when the wire is perpendicular to the lever arm, from one corner of the sign to the other. Thus, T (1.50 m)2 + (0.80 m)2 = 225 N ? m, or T = 132 N. The angle that the wire makes with the horizontal is 90?? – arctan ( 0.80 ) = 62.0??. Thus, the vertical component of 1.50 the force that the pivot exerts is (300 N) ÔÇô(132 N) sin 62.0?? = 183 N and the horizontal force is (132 N) cos 62.0?? = 62 N , for a magnitude of 193 N and an angle of 71?? above the horizontal.

-4 2 m) 0 b) l 1w F? = AY = AY l ?w (2.1?ù1011Pa)(?(2.0?ù10-2m)2) 0.10?ù10-3m = = 3.1?ù106 N, 0.42 2.0 ?ù10-2 m

11.65: a) The tension in the horizontal part of the wire will be 240 N. Taking torques about the center of the disk, (240 N)(0.250 m) – w(1.00 m)) = 0, or w = 60 N. b) Balancing torques about the center of the disk in this case, (240 N) (0.250 m) – ((60 N)(1.00 m) + (20 N)(2.00 m)) cos ? = 0, so ? = 53.1?? .

11.66: a) Taking torques about the right end of the stick, the friction force is half the weight of the stick, f = w ? Taking torques about the point where the cord is attached to 2

the wall (the tension in the cord and the friction force exert no torque about this point),and noting that the moment arm of the normal force is l tan ?, n tan ? = w ? Then, f = tan ? < 0.40, so ? < arctan (0.40) = 22??. 2n

b) Taking torques as in part (a), and denoting the length of the meter stick as l, ll 22 In terms of the coefficient of friction ?Á , s

f l + (l – x) 3l – 2x n l + x l + 2x s 2

Solving for x, l 3tan? – ?Á 2 ?Á + tan ? s c) In the above expression, setting x = 10 cm and solving for ?Á gives s

center of mass is displaced horizontally by a distance (0.625 m – 0.25 m) sin 45?? and the horizontal distance to the point where the upper person is lifting is (1.25 m) sin 45?? , and sotheupperliftswithaforceofw0.375sin45??=(0.300)w=588N.Thepersononthe 1.25 sin 45?? bottom lifts with a force that is the difference between this force and the weight, 1.37 kN. The person above is lifting less.

(a) ?? = 0 Elbow F (3.80 cm) = (15.0 N)(15.0 cm) B F = 59.2 N B

(b) ?? = 0 E F (3.80 cm) = (15.0 N)(15.0 cm) + (80.0 N)(33.0 cm) B

?? = 0, axis at elbow wL – (T sin ? )D = 0 h hD sin ? = so w = T h2 + D2 L h2 + D2 hD w =T max max L h2 + D2 dw T h ? D2 ? ? 2 + 2 ? b) max = max 1 – ; the derivative is positive dD L h2 + D2 ? h D ? c) The result of part (b) shows that w increases when D increases. max

11.70: ?? (about right end) = 0 : 2A(3.6 m) = (90.0 N)(1.8 m) + (1500 N)(0.50 m) A = 130 N = B ?F = 0 : A + B + C + D = 1590 N Use A = B = 130 N and C = D C = D = 670 N By Newton’s third law of motion, the forces A, B, C, and D on the table are the same as the forces the table exerts on the floor.

V and H are the vertical and horizontal forces each wall exerts on the roof. 2V = w so V = w 2 Apply ?? = 0 to one half of the roof, with the axis along the line where the two (w2)(L2)(cos35.0??)+HLsin35.0??-VLcos35.0??=0 L divides out, and use V = w 2 H sin 35.0?? = 1 wcos 35.0?? 4

w H = = 7140 N 4 tan 35.0?? By Newton’s 3rd law, the roof exerts a horizontal, outward force on the wall. For torque about an axis at the lower end of the wall, at the ground, this force has a larger moment arm and hence larger torque the taller the walls. b)

r 11.72: a) Take torques about the upper corner of the curb. The force F acts at a perpendicular distance R – h and the weight acts at a perpendicular distance R2-(R-h)2= 2Rh-h2.Settingthetorquesequalfortheminimumnecessaryforce, 2Rh – h2 R-h r b) The torque due to gravity is the same, but the force F acts at a perpendicular distance 2R – h, so the minimum force is (mg) 2Rh – hv /2RÔÇôh. c) Less force is required when the force is applied at the top of the wheel.

11.73: a) There are several ways to find the tension. Taking torques about point B (the force of the hinge at A is given as being vertical, and exerts no torque about B), the tension acts at distance r = (4.00 m)2 + (2.00 m)2 = 4.47 m and at an angle of ? 2.00 ? ? = 30?? + arctan ? ? = 56.6??. Setting ? 4.00 ? Tr sin ? = (500 N) (2.00 m) and solving for T gives T = 268 N . b) The hinge at A is given as exerting no horizontal force, so taking torques about point D, the lever arm for the verticalforceatpointBis(2.00m)+(4.00m)tan30.0??=4.31m,sothehorizontalforce (500 N) (2.00 m) at B is = 232 N. Using the result of part (a), 4.31 m however, (268 N) cos 30.0?? = 232 N In fact, finding the horizontal force at B first simplifiesthecalculationofthetensionslightly.c)(500N)-(268N)sin30.0??=366N. Equivalently, the result of part (b) could be used, taking torques about point C, to get the same result.

block. The center of gravity of this combination is then 3L 4 from the right edge of the b) Take the two-block combination from part (a), and place it on the third block such that the overhang of 3L 4 is from the right edge of the third block; that is, the center of gravity of the first two blocks is above the right edge of the third block. The center of mass of the three-block combination, measured from the right end of the bottom block, is Similarly, placing this three-block combination with its center of gravity over the right edge of the fourth block allows an extra overhang of L 8, for a total of 25L 24. c) As the result of part (b) shows, with only four blocks, the overhang can be larger than the 11.75: a)

F = 2w = 1.47 N B sin ? = R 2R so ? = 30?? ? = 0, axis at P F(2Rcos?)-wR=0 C

mg F = = 0.424 N C 2cos30?? F = F = 0.424 N AC b) Consider the forces on the bottom marble. The horizontal forces must sum to zero, so F = n sin ? A

F n = A = 0.848 N sin 30?? Could use instead that the vertical forces sum to zero F – mg – n cos ? = 0 B

as an axis and taking counterclockwise rotation as positive: ?L?L? F L sin – F cos – w sin = 0 wire c 22222 where? is the angle between the beams, F is the force exerted by the cross bar, and w is c

the weight of one beam. The length drops out, and all other quantities except F are c

known, so F sin ? – 1 w sin ? ? F = wire 2 2 2 = (2F – w) tan 1 cos ? wire c 2 22

Therefore 53?? F = 260 tan = 130 N 2 b) The cross bar is under compression, as can be seen by imagining the behavior of the two beams if the cross bar were removed. It is the cross bar that holds them apart.

c) The upward pull of the wire on each beam is balanced by the downward pull of gravity, due to the symmentry of the arrangement. The hinge therefore exerts no vertical force. It must, however, balance the outward push of the cross bar: 130 N horizontally to the left for the right-hand beam and 130 N to the right for the left-hand beam. Again, it’s instructive to visualize what the beams would do if the hinge were removed.

11.77: a) The angle at which the bale would slip is that for which f=?Á?=?Áwcos?=wsin?,or?=arctan(?Á)=31.0??.Theangleatwhichthebale ss s would tip is that for which the center of gravity is over the lower contact point, or arctan ( = 26.6??, or 27?? to two figures. The bale tips before it slips. b) The angle for 0.25 m) 0.50 m) tipping is unchanged, but the angle for slipping is arctan (0.40) = 21.8??, or 22?? to two figures. The bale now slips before it tips.

11.78: a) F = f = ?Á ? = ?Á mg = (0.35)(30.0 kg)(9.80 m s2 ) = 103 N kk b) With respect to the forward edge of the bale, the lever arm of the weight is 0.250 m = 0.125 m and the lever arm h of the applied force is then h 2

respect to this point, the weight exerts a counterclockwise torque and the applied force and the force of wheel A both exert clockwise torques. Balancing torques, F(2.00m)+(F)(1.60m)=(950N)(1.00m).Using A

F = ?Á w = 494 N, F = 80 N, and F = w – F = 870 N. b) Again taking torques about the kABA point where wheel B is in contact with the tract, and using F = 494 N as in part (a), (494 N) h = (950 N)(1.00N), so h = 1.92 m.

11.80: a) The torque exerted by the cable about the left end is TL sin ? . For any angle ?,sin(180??-?)=sin?,sothetensionTwillbethesameforeitherangle.Thehorizontal component of the force that the pivot exerts on the boom will be T cos? or Tcos (180?? – ?) = -T cos? . b) From the result of part (a), T ? 1 , and this sin ? becomes infinite as ? ? 0 or ? 180??. Also, c), the tension is a minimum when sin ? is a maximum, or ? = 90??, a vertical string. d) There are no other horizontal forces, so for the boom to be in equilibrium, the pivot exerts zero horizontal force on the boom.

11.81: a) Taking torques about the contact point on the ground, T(7.0m)sin?=w(4.5m)sin?,soT=(0.64)w=3664N.Thegroundexertsavertical force on the pole, of magnitude w – T = 2052 N . b) The factor of sin ? appears in both terms of the equation representing the balancing of torques, and cancels.

0 0 11.83: a) At the bottom of the path the wire exerts a force equal in magnitude to the centripetal acceleration plus the weight, From Eq. (11.10), the elongation is (1.07 ?ù103 N)(0.50 m) (0.7?ù1011Pa)(0.014?ù10-4m2)

b) The ratio of the added force to the elongation, found from taking the slope of the graph, doing a least-squares fit to the linear part of the data, or from a casual glance at the datagivesF=2.00?ù104Nm.FromEq.(11.10), l

F l (3.50 m) lA (?(0.35?ù10-3m)2) c) The total force at the proportional limit is 20.0 N + 60 N = 80 N, and the stress at (80 N) 8 ? (0.35?ù10 3 m)2

11.85: a) For the same stress, the tension in wire B must be two times in wire A, and so the weight must be suspended at a distance (2 3)(1.05 m) = 0.70 m from wire A..

b) The product Y A for wire B is (4 3) that of wire B, so for the same strain, the tension in wire B must be (4 3) that in wire A, and the weight must be 0.45 m from wire B.

11.86: a) Solving Eq. (11.10) for l and using the weight for F, Fl (1900 N)(15.0 m) – YA (2.0?ù1011Pa)(8.00?ù10-4m2) b) From Example 5.21, the force that each car exerts on the cable is F = m?2l = w w2l , and so 0g0

(11.10), with l = l and F = F , 1212 ?AY? ?(1.00cm2)(21?ù1010Pa)? L2 = L1 ? ? = (1.40 m)? ? = 1.63 m. 22 ?AY? ?(2.00cm2)(9?ù1010Pa)? 11 b)Fornickel,F=4.00?ù108Paandforbrass,F=2.00?ù108Pa.c)Fornickel, A1 A2 8

21?ù1010 Pa 9?ù1010 Pa

? l? – 11.88: a)F =YA? ? =(1.4?ù1010Pa)(3.0?ù104m2)(0.010)=4.2?ù104N. max ? l0 ?max

b) Neglect the mass of the shins (actually the lower legs and feet) compared to the rest of the body. This allows the approximation that the compressive stress in the shin bones is uniform. The maximum height will be that for which the force exerted on each lower leg by the ground is F found in part (a), minus the person’s weight. The impulse max that the ground exerts is J=(4.2?ù104N-(70kg)(9.80ms2))(0.030s)=1.2?ù103kg?ms.Thespeedatthe ground is 2gh, so 2J = m 2gh and solving for h,

1 ? 2J ?2 h = ? ? = 64 m, 2g ? m ? 11.89: a) Two times as much, 0.36 mm, b) One-fourth (which is (1 2) 2 ) as much, 0.045 mm.c) The Young’s modulus for copper is approximately one-half that for steel, so the wire would stretch about twice as much. (0.18 mm) 20?ù10 Pa = 0.33 mm. 10 11?ù1010 Pa

11.90: Solving Eq. (11.14) for V , mg V = -kV P = -kV 00 A

(110?ù10-11Pa-1)(250L)(1420kg)(9.80ms2) =- ?(0.150 m)2 The minus sign indicates that this is the volume by which the original hooch has shrunk, and is the extra volume that can be stored.

intersectionoftheredplaneandthebarinFigure(11.52))isA/cos?,sothe b) The tangential component of the force is F sin ?, so the shear stress is c) cos2? is a maximum when cos ? = 1, or ? = 0. d) The shear stress can be expressedas(F2A)sin(2?),whichismaximizedwhen 90?? sin (2?) = 1, or ? = = 45??. Differentiation of the original expression with respect 2 to ? and setting the derivative equal to zero gives the same result.

11.92: a) Taking torques about the pivot, the tension T in the cable is related to mg the weight by T sin ? l = mgl 2, so T = . The horizontal component of the 00 2 sin ? force that the cable exerts on the rod, and hence the horizontal component of the mg mg force that the pivot exerts on the rod, is cot ? and the stress is cot ?. 2 2A b) l F mgl cot ? AY 2AY c) In terms of the density and length, (m A) = ? l , so the stress is 0

( ?l g 2) cot ? and the change in length is ( ?l 2 g 2? ) cot ?. d) Using the numerical 00 values, the stress is 1.4?ù105 Pa and the change in length is 2.2?ù10-6 m. e) The stress is proportional to the length and the change in length is proportional to the square of the length, and so the quantities change by factors of 2 and 4.

tip when F = ? = 750 ?, and would slip when F = (?Á )(1500 ?) = 600 ?, so (1500 )(0.90 m) (1.80 m) s the bookcase slides before tipping. b) If F is vertical, there will be no net horizontal force and the bookcase could not slide. Again taking torques about the left edge of the left leg, the force necessary to tip the case is ? = 13.5 kN. (1500 )(0.90 m) (0.10 m)

c) To slide, the friction force is f = ?Á ( w + F cos ? ), and setting this equal s

ground, the lever arm of the applied force is h and the lever arm of both the 2

weight and the normal force is h tan ?, and so F h = (n – w)h tan ?. Taking torques 2

about the upper point (where the rope is attached to the post), f h = F h . Using 2

f ? ?Á n and solving for F, s ? ?-1 -1 1 1 ?1 1 ? F ? 2w? – ? = 2(400 nN )? – ? = 400 nN , ? ?Á tan ?? ? 0.30 tan 36.9?? ? s

b) The above relations between F, n and f become 3 ?2 F h = (n – w)h tan , f = F , 55 and eliminating f and n and solving for F gives -1 ?25 35? F ? w? – ? , ? ?Á tan?? s

the cable is attached a distance x to the right of the pivot. The sine of the angle between the lever arm and the cable is then h h2 + ((L 2) – x)2 , and the tension is ? hx ? T ? ? = w L 2, ? h + ((L 2) – x) ?? 22 ? where w is the total load (the exact value of w and the position of the center of gravity do not matter for the purposes of this problem). The minimum tension will occur when the term in square brackets is a maximum; differentiating and setting the derviative equal to zero gives a maximum, and hence a minimum tension, at x = (h2 L) + (L 2). However, if x > L, which occurs if h > L 2, the cable must min min be attached at L, the furthest point to the right.

11.96: The geometry of the 3-4-5 right triangle simplifies some of the intermediate algebra. Denote the forces on the ends of the ladders by F and F (left and right). The contact forces at the ground will be vertical, since LR the floor is assumed to be frictionless. a) Taking torques about the right end, F (5.00 m) = (480 N)(3.40 m) + (360 N)(0.90 m), so F = 391 N. F may be found in a L LR similar manner, or from F = 840 N – F = 449 N. b) The tension in the rope may be RL found by finding the torque on each ladder, using the point A as the origin. The lever arm of the rope is 1.50 m. For the left ladder, T (1.50 m) = F (3.20 m) – (480 N)(1.60 m), so T = 322.1 N (322 N to three figures). As a L

check, using the torques on the right ladder, T (1.50 m) = F (1.80 m) – (360 N)(0.90 m) gives the same result. c) The horizontal R

component of the force at A must be equal to the tension found in part (b). The vertical force must be equal in magnitude to the difference between the weight of each ladder and the force on the bottom of each ladder, 480 N-391 N = 449 N-360 N = 89 N. The magnitude of the force at A is then d) The easiest way to do this is to see that the added load will be distributed at the floor in such a way that F?=F+(0.36)(800N)=679N,andF?=F+(0.64)(800N)=961N.Usingthese LL RR forces in the form for the tension found in part (b) gives F?(3.20m)-(480N)(1.60m) F?(1.80m)-(360N)(0.90m) T = L = R = 936.53 N, (1.50 m) (1.50 m) which is 937 N to three figures.

OO volume of the sodium is = k v p. Setting the total volume change equal to Ax (x is ss positive) and using p = F A, Ax = (k V + k V )(F A), OO ss and solving for k gives s

? A2 x ? 1 k = ? – k V ? ? s OO ? F ? Vs

11.98: a) For constant temperature ( T = 0), ( p)V ( V) b) In this situation, ( p)V ? + ?p( V )V ?-1 = 0, ( p) + ?p V = 0, V and ( p)V V

11.99:a)FromEq.(11.10),l= 2-m)=6.62?ù10-4m,or0.66mm (4.50 kg)(9.80 m s )(1.50 ( 20?ù1010 Pa )(5.00?ù10 7 m 2 ) totwofigures.b)(4.50kg)(9.80ms2)(0.0500?ù10-2m)=0.022J.c)Themagnitude F will be vary with distance; the average force is Y A(0.0250 cm l ) = 16.7 N, and so 0